标签:names ota code modify ide lag sum 资料 first
〖模板代码〗
1 void getroot(int x,int fa) 2 { 3 sz[x]=1;mx[x]=0; 4 for(int i=first[x];i;i=e[i].next) 5 { 6 int to=e[i].to; 7 if(to==fa||vis[to])continue; 8 getroot(to,x); 9 sz[x]+=sz[to]; 10 mx[x]=max(mx[x],sz[to]); 11 } 12 mx[x]=max(mx[x],sum-sz[x]); 13 if(mx[root]>mx[x])root=x; 14 } 15 void getdeep(int x,int fa) 16 { 17 d[++tot]=deep[x]; 18 for(int i=first[x];i;i=e[i].next) 19 { 20 int to=e[i].to; 21 if(to==fa||vis[to])continue; 22 deep[to]=deep[x]+e[i].w; 23 getdeep(to,x); 24 } 25 } 26 int calc(int x) 27 { 28 tot=0;getdeep(x,-1); 29 sort(d+1,d+tot+1); 30 int ans=0,i=1,j=tot; 31 while(i<j) 32 { 33 if(d[i]+d[j]<=K)ans+=j-i,i++; 34 else j--; 35 } 36 return ans; 37 } 38 void dfs(int x) 39 { 40 deep[x]=0;vis[x]=true;ans+=calc(x); 41 for(int i=first[x];i;i=e[i].next) 42 { 43 int to=e[i].to; 44 if(vis[to])continue; 45 deep[to]=e[i].w; 46 ans-=calc(to); 47 sum=sz[to];root=0; 48 getroot(to,-1);dfs(root); 49 } 50 } 51 int main() 52 { 53 root=0;sum=n;mx[0]=inf; 54 getroot(1,-1);dfs(root); 55 printf("%d\n",ans); 56 }
〖相关题目〗
1.【poj1741】Tree
题意:给定一棵n个点的有边权的树,给定数字k,求满足x到y距离≤k的无序点对(x,y)的个数。
分析:GXZlegendの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 const int N=1e4+5; 6 const int inf=0x3f3f3f3f; 7 int n,K,cnt,x,y,w,root,sum,ans,tot; 8 int sz[N],mx[N],first[N],deep[N],d[N]; 9 bool vis[N]; 10 struct edge{int to,next,w;}e[N<<1]; 11 int read() 12 { 13 int x=0,f=1;char c=getchar(); 14 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 15 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 16 return x*f; 17 } 18 void ins(int u,int v,int w){e[++cnt]=(edge){v,first[u],w};first[u]=cnt;} 19 void getroot(int x,int fa) 20 { 21 sz[x]=1;mx[x]=0; 22 for(int i=first[x];i;i=e[i].next) 23 { 24 int to=e[i].to; 25 if(to==fa||vis[to])continue; 26 getroot(to,x); 27 sz[x]+=sz[to]; 28 mx[x]=max(mx[x],sz[to]); 29 } 30 mx[x]=max(mx[x],sum-sz[x]); 31 if(mx[root]>mx[x])root=x; 32 } 33 void getdeep(int x,int fa) 34 { 35 d[++tot]=deep[x]; 36 for(int i=first[x];i;i=e[i].next) 37 { 38 int to=e[i].to; 39 if(to==fa||vis[to])continue; 40 deep[to]=deep[x]+e[i].w; 41 getdeep(to,x); 42 } 43 } 44 int calc(int x) 45 { 46 tot=0;getdeep(x,-1); 47 sort(d+1,d+tot+1); 48 int ans=0,i=1,j=tot; 49 while(i<j) 50 { 51 if(d[i]+d[j]<=K)ans+=j-i,i++; 52 else j--; 53 } 54 return ans; 55 } 56 void dfs(int x) 57 { 58 deep[x]=0;vis[x]=true;ans+=calc(x); 59 for(int i=first[x];i;i=e[i].next) 60 { 61 int to=e[i].to; 62 if(vis[to])continue; 63 deep[to]=e[i].w; 64 ans-=calc(to); 65 sum=sz[to];root=0; 66 getroot(to,-1);dfs(root); 67 } 68 } 69 int main() 70 { 71 n=read();K=read(); 72 while(n||K) 73 { 74 cnt=0;ans=0; 75 memset(first,0,sizeof(first)); 76 memset(vis,0,sizeof(vis)); 77 for(int i=1;i<n;i++) 78 { 79 x=read();y=read();w=read(); 80 ins(x,y,w);ins(y,x,w); 81 } 82 root=0;sum=n;mx[0]=inf; 83 getroot(1,-1);dfs(root); 84 printf("%d\n",ans); 85 n=read();K=read(); 86 } 87 return 0; 88 }
2.【bzoj2152】聪聪可可
题意:给定一棵n个点的有边权的树,求满足x到y距离恰好是3的倍数的有序点对(x,y)的个数/有序点对总个数。
分析:点分治裸题
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 const int N=2e4+5; 6 const int inf=0x3f3f3f3f; 7 int n,cnt,x,y,w,root,sum,ans; 8 int t[3],first[N],mx[N],sz[N],deep[N]; 9 bool vis[N]; 10 struct edge{int to,next,w;}e[N<<1]; 11 int read() 12 { 13 int x=0,f=1;char c=getchar(); 14 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 15 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 16 return x*f; 17 } 18 void ins(int u,int v,int w){e[++cnt]=(edge){v,first[u],w};first[u]=cnt;} 19 void getroot(int x,int fa) 20 { 21 sz[x]=1;mx[x]=0; 22 for(int i=first[x];i;i=e[i].next) 23 { 24 int to=e[i].to; 25 if(to==fa||vis[to])continue; 26 getroot(to,x); 27 sz[x]+=sz[to]; 28 mx[x]=max(mx[x],sz[to]); 29 } 30 mx[x]=max(mx[x],sum-sz[x]); 31 if(mx[root]>mx[x])root=x; 32 } 33 void getdeep(int x,int fa) 34 { 35 t[deep[x]]++; 36 for(int i=first[x];i;i=e[i].next) 37 { 38 int to=e[i].to; 39 if(to==fa||vis[to])continue; 40 deep[to]=(deep[x]+e[i].w)%3; 41 getdeep(to,x); 42 } 43 } 44 int calc(int x) 45 { 46 t[0]=t[1]=t[2]=0;getdeep(x,-1); 47 return t[1]*t[2]*2+t[0]*t[0]; 48 } 49 void dfs(int x) 50 { 51 deep[x]=0;vis[x]=true;ans+=calc(x); 52 for(int i=first[x];i;i=e[i].next) 53 { 54 int to=e[i].to; 55 if(vis[to])continue; 56 deep[to]=e[i].w; 57 ans-=calc(to); 58 sum=sz[to];root=0; 59 getroot(to,-1);dfs(root); 60 } 61 } 62 int gcd(int a,int b){return !b?a:gcd(b,a%b);} 63 int main() 64 { 65 n=read(); 66 for(int i=1;i<n;i++) 67 { 68 x=read();y=read();w=read()%3; 69 ins(x,y,w);ins(y,x,w); 70 } 71 root=0;sum=n;mx[0]=inf; 72 getroot(1,-1);dfs(root); 73 int g=gcd(ans,n*n); 74 printf("%d/%d\n",ans/g,n*n/g); 75 return 0; 76 }
3.【bzoj2599】[IOI2011]Race
题意:给一棵树,每条边有权。求一条简单路径,权值和等于K,且边的数量最小。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 const int N=2e5+5; 6 const int inf=0x3f3f3f3f; 7 int n,K,cnt,x,y,w,root,sum,ans,froot; 8 int first[N],sz[N],deep[N],dis[N],t[N*5]; 9 bool vis[N]; 10 struct edge{int to,next,w;}e[N<<1]; 11 int read() 12 { 13 int x=0,f=1;char c=getchar(); 14 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 15 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 16 return x*f; 17 } 18 void ins(int u,int v,int w){e[++cnt]=(edge){v,first[u],w};first[u]=cnt;} 19 void getroot(int x,int fa) 20 { 21 bool flag=true;sz[x]=1; 22 for(int i=first[x];i;i=e[i].next) 23 { 24 int to=e[i].to; 25 if(to==fa||vis[to])continue; 26 getroot(to,x); 27 sz[x]+=sz[to]; 28 if(sz[to]>sum/2)flag=false; 29 } 30 if(sum-sz[x]>sum/2)flag=false; 31 if(flag)root=x,froot=fa; 32 } 33 void getdeep(int x,int fa) 34 { 35 if(dis[x]<=K)ans=min(ans,deep[x]+t[K-dis[x]]); 36 for(int i=first[x];i;i=e[i].next) 37 { 38 int to=e[i].to; 39 if(to==fa||vis[to])continue; 40 dis[to]=dis[x]+e[i].w; 41 deep[to]=deep[x]+1; 42 getdeep(to,x); 43 } 44 } 45 void check(int x,int fa,int flag) 46 { 47 if(dis[x]<=K) 48 { 49 if(flag)t[dis[x]]=min(t[dis[x]],deep[x]); 50 else t[dis[x]]=n; 51 } 52 for(int i=first[x];i;i=e[i].next) 53 { 54 int to=e[i].to; 55 if(to==fa||vis[to])continue; 56 check(to,x,flag); 57 } 58 } 59 void dfs(int x,int s,int fa) 60 { 61 vis[x]=true;dis[x]=deep[x]=0;t[0]=0; 62 for(int i=first[x];i;i=e[i].next) 63 { 64 int to=e[i].to; 65 if(vis[to])continue; 66 deep[to]=1;dis[to]=e[i].w; 67 getdeep(to,x);check(to,x,1); 68 } 69 for(int i=first[x];i;i=e[i].next) 70 { 71 int to=e[i].to; 72 if(vis[to])continue; 73 check(to,x,0); 74 } 75 for(int i=first[x];i;i=e[i].next) 76 { 77 int to=e[i].to; 78 if(vis[to])continue; 79 if(to==fa)sum=s-sz[x]; 80 else sum=sz[to]; 81 root=0;getroot(to,x); 82 dfs(root,sum,froot); 83 } 84 } 85 int main() 86 { 87 n=read();K=read(); 88 for(int i=1;i<=K;i++)t[i]=n; 89 for(int i=1;i<n;i++) 90 { 91 x=read()+1;y=read()+1;w=read(); 92 ins(x,y,w);ins(y,x,w); 93 } 94 root=0;sum=n;ans=n; 95 getroot(1,-1);dfs(root,n,froot); 96 if(ans==n)printf("-1\n"); 97 else printf("%d\n",ans); 98 return 0; 99 }
4.【bzoj3697】采药人的路径
题意:给定一棵n个点的树,边权只有0或1。一条路径是合法的,要求路径上边权为0和1的边数量相等,且路径中有一个点(不包括起点和终点),满足起点到该点和该点到终点的路径上边权为0和1的边数量也相等。求一共可以选择多少种不同的路径。
分析:CQzhangyuの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=1e5+5; 7 const int base=1e5; 8 const int inf=0x3f3f3f3f; 9 int n,cnt,x,y,w,sum,root,d; 10 int first[N],sz[N],mx[N],dep[N]; 11 int s[N<<1],f[N<<1][2],g[N<<1][2]; 12 LL ans; 13 bool vis[N]; 14 struct edge{int to,next,w;}e[N<<1]; 15 int read() 16 { 17 int x=0,f=1;char c=getchar(); 18 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 19 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 20 return x*f; 21 } 22 void ins(int u,int v,int w){e[++cnt]=(edge){v,first[u],w};first[u]=cnt;} 23 void getroot(int x,int fa) 24 { 25 sz[x]=1;mx[x]=0; 26 for(int i=first[x];i;i=e[i].next) 27 { 28 int to=e[i].to; 29 if(to==fa||vis[to])continue; 30 getroot(to,x); 31 sz[x]+=sz[to]; 32 mx[x]=max(mx[x],sz[to]); 33 } 34 mx[x]=max(mx[x],sum-sz[x]); 35 if(mx[root]>mx[x])root=x; 36 } 37 void getdeep(int x,int fa) 38 { 39 d=max(d,max(dep[x],-dep[x])); 40 if(s[dep[x]+base])g[dep[x]+base][1]++; 41 else g[dep[x]+base][0]++; 42 s[dep[x]+base]++; 43 for(int i=first[x];i;i=e[i].next) 44 { 45 int to=e[i].to; 46 if(to==fa||vis[to])continue; 47 dep[to]=dep[x]+e[i].w; 48 getdeep(to,x); 49 } 50 s[dep[x]+base]--; 51 } 52 void dfs(int x) 53 { 54 vis[x]=true;int dd=0; 55 for(int i=first[x];i;i=e[i].next) 56 { 57 int to=e[i].to; 58 if(vis[to])continue; 59 dep[to]=e[i].w;d=0; 60 getdeep(to,x);dd=max(dd,d); 61 for(int j=base-d;j<=base+d;j++) 62 ans+=1ll*f[base*2-j][0]*g[j][1]+1ll*f[base*2-j][1]*g[j][0]+1ll*f[base*2-j][1]*g[j][1]; 63 ans+=1ll*f[base][0]*g[base][0]+g[base][1]; 64 for(int j=base-d;j<=base+d;j++) 65 f[j][0]+=g[j][0],f[j][1]+=g[j][1],g[j][0]=g[j][1]=0; 66 } 67 for(int i=base-dd;i<=base+dd;i++)f[i][0]=f[i][1]=0; 68 for(int i=first[x];i;i=e[i].next) 69 { 70 int to=e[i].to; 71 if(vis[to])continue; 72 root=0;sum=sz[to]; 73 getroot(to,x);dfs(root); 74 } 75 } 76 int main() 77 { 78 n=read(); 79 for(int i=1;i<n;i++) 80 { 81 x=read();y=read();w=read(); 82 w=(w==0)?-1:1;ins(x,y,w);ins(y,x,w); 83 } 84 root=0;sum=n;mx[0]=inf; 85 getroot(1,-1);dfs(root); 86 printf("%lld\n",ans); 87 return 0; 88 }
5.【Codecraft-18 and Codeforces Round #458】E. Palindromes in a Tree
题意:给定n个点的树,每个点有一个a~t的字母,一条路径回文当且仅当路径上的字母的某一个排列回文,求经过每个点的回文路径数。
分析:点分治+状压
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=2e5+5; 7 const int inf=0x3f3f3f3f; 8 int n,cnt,x,y,root,sum; 9 int id[N],first[N],sz[N],mx[N]; 10 bool vis[N]; 11 LL ans[N],mp[(1<<20)+5]; 12 char s[N]; 13 struct edge{int to,next;}e[N*2]; 14 int read() 15 { 16 int x=0,f=1;char c=getchar(); 17 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 18 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 19 return x*f; 20 } 21 int max(int a,int b){return a>b?a:b;} 22 void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;} 23 void getroot(int x,int fa) 24 { 25 sz[x]=1;mx[x]=0; 26 for(int i=first[x];i;i=e[i].next) 27 { 28 int to=e[i].to; 29 if(to==fa||vis[to])continue; 30 getroot(to,x); 31 sz[x]+=sz[to]; 32 mx[x]=max(mx[x],sz[to]); 33 } 34 mx[x]=max(mx[x],sum-sz[x]); 35 if(mx[root]>mx[x])root=x; 36 } 37 void getway(int x,int fa,int val,int k) 38 { 39 val^=id[x];mp[val]+=k; 40 for(int i=first[x];i;i=e[i].next) 41 { 42 int to=e[i].to; 43 if(to==fa||vis[to])continue; 44 getway(to,x,val,k); 45 } 46 } 47 LL check(int x,int fa,int val) 48 { 49 val^=id[x];LL num=mp[val]; 50 for(int i=0;i<20;i++)num+=mp[(1<<i)^val]; 51 for(int i=first[x];i;i=e[i].next) 52 { 53 int to=e[i].to; 54 if(to==fa||vis[to])continue; 55 num+=check(to,x,val); 56 } 57 ans[x]+=num;return num; 58 } 59 void solve(int x) 60 { 61 vis[x]=true;getway(x,-1,0,1); 62 LL num=mp[0]; 63 for(int i=0;i<20;i++)num+=mp[1<<i]; 64 for(int i=first[x];i;i=e[i].next) 65 { 66 int to=e[i].to; 67 if(vis[to])continue; 68 getway(to,x,id[x],-1); 69 num+=check(to,x,0); 70 getway(to,x,id[x],1); 71 } 72 ans[x]+=num/2;getway(x,-1,0,-1); 73 for(int i=first[x];i;i=e[i].next) 74 { 75 int to=e[i].to; 76 if(vis[to])continue; 77 sum=sz[to];root=0; 78 getroot(to,-1);solve(root); 79 } 80 } 81 int main() 82 { 83 n=read(); 84 for(int i=1;i<n;i++)x=read(),y=read(),ins(x,y),ins(y,x); 85 scanf("%s",s+1); 86 for(int i=1;i<=n;i++)id[i]=1<<(s[i]-‘a‘); 87 sum=n;mx[0]=inf;getroot(1,-1);solve(root); 88 for(int i=1;i<=n;i++)printf("%lld ",ans[i]+1); 89 return 0; 90 }
〖相关资料〗
〖模板代码〗
1 struct LCT 2 { 3 int c[N][2],fa[N],sum[N],val[N]; 4 bool rev[N]; 5 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 6 void update(int x){sum[x]=sum[c[x][0]]+sum[c[x][1]]+val[x];} 7 void flip(int x){swap(c[x][0],c[x][1]);rev[x]^=1;} 8 void down(int x){if(rev[x])flip(c[x][0]),flip(c[x][1]),rev[x]=0;} 9 void rotate(int x) 10 { 11 int y=fa[x],z=fa[y],l,r; 12 if(c[y][0]==x)l=0;else l=1;r=l^1; 13 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 14 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 15 c[y][l]=c[x][r];c[x][r]=y; 16 update(y);update(x); 17 } 18 void relax(int x){if(!isroot(x))relax(fa[x]);down(x);} 19 void splay(int x) 20 { 21 relax(x); 22 while(!isroot(x)) 23 { 24 int y=fa[x],z=fa[y]; 25 if(!isroot(y)) 26 { 27 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 28 else rotate(y); 29 } 30 rotate(x); 31 } 32 } 33 void access(int x){int t=0;while(x)splay(x),c[x][1]=t,update(x),t=x,x=fa[x];} 34 void makeroot(int x){access(x);splay(x);flip(x);} 35 bool connected(int x,int y) 36 { 37 if(x==y)return true; 38 makeroot(x);access(y);splay(y);return fa[x]!=0; 39 } 40 void modify(int x,int v){splay(x);val[x]=v;update(x);} 41 void link(int x,int y){makeroot(x);fa[x]=y;} 42 void cut(int x,int y){makeroot(x);access(y);splay(y);c[y][0]=fa[x]=0;update(y);} 43 int query(int x,int y){makeroot(x);access(y);splay(y);return sum[y];} 44 }T;
〖相关题目〗
1.【bzoj3282】Tree
题意:给定n个点以及每个点的权值,处理m个操作。0:询问从x到y的路径上的点的权值的xor和。保证x到y是联通的。1:连接x到y,若x到y已经联通则无需连接。2:删除边(x,y),不保证边(x,y)存在。3:将点X上的权值变成Y。
分析:Link-Cut Tree裸题
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define LL long long 5 using namespace std; 6 const int N=3e5+5; 7 int n,m,op,x,y; 8 int read() 9 { 10 int x=0,f=1;char c=getchar(); 11 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 12 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 13 return x*f; 14 } 15 struct LCT 16 { 17 int c[N][2],fa[N],sum[N],val[N]; 18 bool rev[N]; 19 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 20 void update(int x){sum[x]=sum[c[x][0]]^sum[c[x][1]]^val[x];} 21 void flip(int x){swap(c[x][0],c[x][1]);rev[x]^=1;} 22 void down(int x){if(rev[x])flip(c[x][0]),flip(c[x][1]),rev[x]=0;} 23 void rotate(int x) 24 { 25 int y=fa[x],z=fa[y],l,r; 26 if(c[y][0]==x)l=0;else l=1;r=l^1; 27 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 28 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 29 c[y][l]=c[x][r];c[x][r]=y; 30 update(y);update(x); 31 } 32 void relax(int x){if(!isroot(x))relax(fa[x]);down(x);} 33 void splay(int x) 34 { 35 relax(x); 36 while(!isroot(x)) 37 { 38 int y=fa[x],z=fa[y]; 39 if(!isroot(y)) 40 { 41 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 42 else rotate(y); 43 } 44 rotate(x); 45 } 46 } 47 void access(int x){int t=0;while(x){splay(x);c[x][1]=t;update(x);t=x;x=fa[x];}} 48 void makeroot(int x){access(x);splay(x);flip(x);} 49 void modify(int x,int v){splay(x);val[x]=v;update(x);} 50 void link(int x,int y){makeroot(x);fa[x]=y;} 51 void cut(int x,int y){makeroot(x);access(y);splay(y);c[y][0]=fa[x]=0;update(y);} 52 int query(int x,int y){makeroot(x);access(y);splay(y);return sum[y];} 53 }T; 54 int main() 55 { 56 n=read();m=read(); 57 for(int i=1;i<=n;i++)T.val[i]=T.sum[i]=read(); 58 while(m--) 59 { 60 op=read();x=read();y=read(); 61 if(op==0)printf("%d\n",T.query(x,y)); 62 if(op==1)T.link(x,y); 63 if(op==2)T.cut(x,y); 64 if(op==3)T.modify(x,y); 65 } 66 return 0; 67 }
2.【bzoj1180】[CROATIAN2009]OTOCI
题意:给定n个点以及每个点的权值,处理m个操作。详见题目。
分析:Link-Cut Tree裸题
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define LL long long 5 using namespace std; 6 const int N=3e4+5; 7 int n,m,x,y; 8 char s[10]; 9 int read() 10 { 11 int x=0,f=1;char c=getchar(); 12 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 13 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 14 return x*f; 15 } 16 struct LCT 17 { 18 int c[N][2],fa[N],sum[N],val[N]; 19 bool rev[N]; 20 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 21 void update(int x){sum[x]=sum[c[x][0]]+sum[c[x][1]]+val[x];} 22 void flip(int x){swap(c[x][0],c[x][1]);rev[x]^=1;} 23 void down(int x){if(rev[x])flip(c[x][0]),flip(c[x][1]),rev[x]=0;} 24 void rotate(int x) 25 { 26 int y=fa[x],z=fa[y],l,r; 27 if(c[y][0]==x)l=0;else l=1;r=l^1; 28 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 29 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 30 c[y][l]=c[x][r];c[x][r]=y; 31 update(y);update(x); 32 } 33 void relax(int x){if(!isroot(x))relax(fa[x]);down(x);} 34 void splay(int x) 35 { 36 relax(x); 37 while(!isroot(x)) 38 { 39 int y=fa[x],z=fa[y]; 40 if(!isroot(y)) 41 { 42 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 43 else rotate(y); 44 } 45 rotate(x); 46 } 47 } 48 void access(int x){int t=0;while(x)splay(x),c[x][1]=t,update(x),t=x,x=fa[x];} 49 void makeroot(int x){access(x);splay(x);flip(x);} 50 bool connected(int x,int y) 51 { 52 if(x==y)return true; 53 makeroot(x);access(y);splay(y);return fa[x]!=0; 54 } 55 void modify(int x,int v){splay(x);val[x]=v;update(x);} 56 void link(int x,int y){makeroot(x);fa[x]=y;} 57 void cut(int x,int y){makeroot(x);access(y);splay(y);c[y][0]=fa[x]=0;update(y);} 58 int query(int x,int y){makeroot(x);access(y);splay(y);return sum[y];} 59 }T; 60 int main() 61 { 62 n=read(); 63 for(int i=1;i<=n;i++)T.sum[i]=T.val[i]=read(); 64 m=read(); 65 while(m--) 66 { 67 scanf("%s",s);x=read();y=read(); 68 if(s[0]==‘b‘) 69 { 70 if(T.connected(x,y))printf("no\n"); 71 else printf("yes\n"),T.link(x,y); 72 } 73 if(s[0]==‘p‘)T.modify(x,y); 74 if(s[0]==‘e‘) 75 { 76 if(T.connected(x,y))printf("%d\n",T.query(x,y)); 77 else printf("impossible\n"); 78 } 79 } 80 return 0; 81 }
3.【bzoj2759】一个动态树好题
题意:有n个x[1..n]和n个等式组成的同余方程组:x[i]=k[i]*x[p[i]]+b[i] mod 10007。要应付Q个事务,每个是两种情况之一:一、询问当前x[a]的解,无解输出-1,多解输出-2,否则输出x[a];二、修改一个等式 C a k[a] p[a] b[a]。
分析:PoPoQQQの博客
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define LL long long 5 using namespace std; 6 const int N=3e4+5; 7 const int mod=1e4+7; 8 int n,m,id,k,p,b; 9 char op[2]; 10 int read() 11 { 12 int x=0,f=1;char c=getchar(); 13 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 14 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 15 return x*f; 16 } 17 struct data 18 { 19 int k,b; 20 data(){k=1;b=0;} 21 int calc(int x){return (k*x+b)%mod;} 22 }; 23 data operator + (data a,data b) 24 { 25 data c; 26 c.k=a.k*b.k%mod; 27 c.b=(b.k*a.b%mod+b.b)%mod; 28 return c; 29 } 30 int exgcd(int a,int b,int& x,int& y) 31 { 32 if(!b){x=1;y=0;return a;} 33 int ans=exgcd(b,a%b,x,y); 34 int tmp=x;x=y;y=tmp-a/b*y; 35 return ans; 36 } 37 struct LCT 38 { 39 int c[N][2],fa[N],sfa[N]; 40 bool ins[N],vis[N]; 41 data val[N],sum[N]; 42 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 43 void update(int x){sum[x]=sum[c[x][0]]+val[x]+sum[c[x][1]];} 44 void dfs(int x) 45 { 46 ins[x]=vis[x]=true;int to=fa[x]; 47 if(ins[to])fa[x]=0,sfa[x]=to; 48 if(!vis[to])dfs(to);ins[x]=false; 49 } 50 void init(int n) 51 { 52 int k,b; 53 for(int i=1;i<=n;i++) 54 { 55 k=read();fa[i]=read();b=read(); 56 data tmp;tmp.k=k;tmp.b=b; 57 val[i]=sum[i]=tmp; 58 } 59 for(int i=1;i<=n;i++)if(!vis[i])dfs(i); 60 } 61 void rotate(int x) 62 { 63 int y=fa[x],z=fa[y],l,r; 64 if(c[y][0]==x)l=0;else l=1;r=l^1; 65 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 66 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 67 c[y][l]=c[x][r];c[x][r]=y; 68 update(y);update(x); 69 } 70 void splay(int x) 71 { 72 while(!isroot(x)) 73 { 74 int y=fa[x],z=fa[y]; 75 if(!isroot(y)) 76 { 77 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 78 else rotate(y); 79 } 80 rotate(x); 81 } 82 } 83 void access(int x){int t=0;while(x)splay(x),c[x][1]=t,update(x),t=x,x=fa[x];} 84 int findroot(int x) 85 { 86 access(x);splay(x); 87 int t=x;while(c[t][0])t=c[t][0]; 88 splay(t);return t; 89 } 90 int query(int x) 91 { 92 access(x);splay(x);data v1=sum[x]; 93 int rt=findroot(x),sf=sfa[rt]; 94 access(sf);splay(sf);data v2=sum[sf]; 95 if(v2.k==0)return v1.calc(v2.b); 96 if(v2.k==1)return v2.b?-1:-2; 97 int xx,yy;exgcd(v2.k-1,mod,xx,yy); 98 // return v1.calc((mod-xx)%mod*v2.b%mod); 99 return v1.calc((xx%mod+mod)%mod*(((mod-v2.b)%mod+mod)%mod)%mod); 100 } 101 void cut(int x){access(x);splay(x);fa[c[x][0]]=0;c[x][0]=0;update(x);} 102 void link(int x,int f){access(x);splay(x);fa[x]=f;} 103 bool oncircle(int x,int rt) 104 { 105 int sf=sfa[rt]; 106 if(x==sf)return true; 107 access(sf);splay(sf);splay(x); 108 return !isroot(sf); 109 } 110 void change(int x,int f,int k,int b) 111 { 112 access(x);splay(x); 113 data tmp;tmp.k=k;tmp.b=b; 114 val[x]=tmp;update(x); 115 int rt=findroot(x); 116 if(x==rt) 117 { 118 int rf=findroot(f); 119 if(rt==rf)sfa[x]=f; 120 else sfa[x]=0,link(x,f); 121 } 122 else 123 { 124 if(oncircle(x,rt)) 125 { 126 cut(x);link(rt,sfa[rt]);sfa[rt]=0; 127 int rf=findroot(f); 128 if(x==rf)sfa[x]=f; 129 else link(x,f); 130 } 131 else 132 { 133 cut(x);int rf=findroot(f); 134 if(x==rf)sfa[x]=f; 135 else link(x,f); 136 } 137 } 138 } 139 }lct; 140 int main() 141 { 142 n=read();lct.init(n); 143 m=read(); 144 while(m--) 145 { 146 scanf("%s",op);id=read(); 147 if(op[0]==‘A‘)printf("%d\n",lct.query(id)); 148 else k=read(),p=read(),b=read(),lct.change(id,p,k,b); 149 } 150 return 0; 151 }
4.【bzoj4530】[Bjoi2014]大融合
题意:给定n个点,逐条加边,一条边的负载就是它所在的当前能够联通的树上路过它的简单路径的数量。随着边的添加,动态的回答对于某些边的负载的询问。
分析:GXZlegendの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=1e5+5; 7 int n,m,x,y; 8 char op[2]; 9 int read() 10 { 11 int x=0,f=1;char c=getchar(); 12 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 13 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 14 return x*f; 15 } 16 struct LCT 17 { 18 int c[N][2],fa[N],sum[N],val[N]; 19 bool rev[N]; 20 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 21 void update(int x){sum[x]=sum[c[x][0]]+sum[c[x][1]]+val[x]+1;} 22 void flip(int x){swap(c[x][0],c[x][1]);rev[x]^=1;} 23 void down(int x){if(rev[x])flip(c[x][0]),flip(c[x][1]),rev[x]=0;} 24 void rotate(int x) 25 { 26 int y=fa[x],z=fa[y],l,r; 27 if(c[y][0]==x)l=0;else l=1;r=l^1; 28 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 29 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 30 c[y][l]=c[x][r];c[x][r]=y; 31 update(y);update(x); 32 } 33 void relax(int x){if(!isroot(x))relax(fa[x]);down(x);} 34 void splay(int x) 35 { 36 relax(x); 37 while(!isroot(x)) 38 { 39 int y=fa[x],z=fa[y]; 40 if(!isroot(y)) 41 { 42 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 43 else rotate(y); 44 } 45 rotate(x); 46 } 47 } 48 void access(int x) 49 { 50 int t=0; 51 while(x) 52 { 53 splay(x);val[x]+=sum[c[x][1]]-sum[t]; 54 c[x][1]=t;update(x);t=x;x=fa[x]; 55 } 56 } 57 void makeroot(int x){access(x);splay(x);flip(x);} 58 void link(int x,int y) 59 { 60 makeroot(x);makeroot(y); 61 fa[x]=y;val[y]+=sum[x];update(y); 62 } 63 }T; 64 int main() 65 { 66 n=read();m=read(); 67 for(int i=1;i<=n;i++)T.sum[i]=1; 68 while(m--) 69 { 70 scanf("%s",op); 71 x=read();y=read(); 72 if(op[0]==‘A‘)T.link(x,y); 73 else 74 { 75 T.makeroot(x);T.makeroot(y); 76 printf("%lld\n",1ll*T.sum[x]*(T.sum[y]-T.sum[x])); 77 } 78 } 79 return 0; 80 }
5.【bzoj 3779】重组病毒
题意:见原题。
分析:Zsnuoの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 #define lc(x) x<<1 6 #define rc(x) x<<1|1 7 using namespace std; 8 const int N=1e5+5; 9 int n,m,x,y,cnt,dfn,rt=1,first[N]; 10 int deep[N],in[N],out[N],fa[N][18]; 11 char op[15]; 12 int read() 13 { 14 int x=0,f=1;char c=getchar(); 15 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 16 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 17 return x*f; 18 } 19 struct edge{int to,next;}e[N<<1]; 20 void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;} 21 struct SGT 22 { 23 int l[N*4],r[N*4]; 24 LL sum[N*4],tag[N*4]; 25 void build(int x,int L,int R) 26 { 27 l[x]=L;r[x]=R;if(L==R)return; 28 int mid=(L+R)>>1; 29 build(lc(x),L,mid);build(rc(x),mid+1,R); 30 } 31 void up(int x){sum[x]=sum[lc(x)]+sum[rc(x)];} 32 void qadd(int x,LL w){tag[x]+=w;sum[x]+=w*(r[x]-l[x]+1);} 33 void down(int x) 34 { 35 if(!tag[x])return; 36 qadd(lc(x),tag[x]);qadd(rc(x),tag[x]); 37 tag[x]=0; 38 } 39 void add(int x,int L,int R,LL w) 40 { 41 if(L<=l[x]&&r[x]<=R){qadd(x,w);return;} 42 down(x);int mid=(l[x]+r[x])>>1; 43 if(L<=mid)add(lc(x),L,R,w); 44 if(R>mid)add(rc(x),L,R,w); 45 up(x); 46 } 47 LL query(int x,int L,int R) 48 { 49 if(L<=l[x]&&r[x]<=R)return sum[x]; 50 down(x);LL ans=0; 51 int mid=(l[x]+r[x])>>1; 52 if(L<=mid)ans+=query(lc(x),L,R); 53 if(R>mid)ans+=query(rc(x),L,R); 54 return ans; 55 } 56 }sgt; 57 int find(int x,int y) 58 { 59 int d=deep[y]-deep[x]-1; 60 for(int i=16;i>=0;i--) 61 if(d&(1<<i))y=fa[y][i]; 62 return y; 63 } 64 void addson(int x,LL w) 65 { 66 if(x==rt)sgt.add(1,1,n,w); 67 else if(in[rt]>=in[x]&&out[rt]<=out[x]) 68 { 69 int t=find(x,rt); 70 if(in[t]>1)sgt.add(1,1,in[t]-1,w); 71 if(out[t]<n)sgt.add(1,out[t]+1,n,w); 72 } 73 else sgt.add(1,in[x],out[x],w); 74 } 75 struct LCT 76 { 77 int c[N][2],fa[N],in[N],out[N]; 78 bool rev[N]; 79 bool isroot(int x){return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;} 80 void flip(int x){swap(c[x][0],c[x][1]);rev[x]^=1;} 81 void down(int x){if(rev[x])flip(c[x][0]),flip(c[x][1]),rev[x]=0;} 82 void rotate(int x) 83 { 84 int y=fa[x],z=fa[y],l,r; 85 if(c[y][0]==x)l=0;else l=1;r=l^1; 86 if(!isroot(y)){if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;} 87 fa[x]=z;fa[y]=x;fa[c[x][r]]=y; 88 c[y][l]=c[x][r];c[x][r]=y; 89 } 90 void relax(int x){if(!isroot(x))relax(fa[x]);down(x);} 91 void splay(int x) 92 { 93 relax(x); 94 while(!isroot(x)) 95 { 96 int y=fa[x],z=fa[y]; 97 if(!isroot(y)) 98 { 99 if((c[y][0]==x)^(c[z][0]==y))rotate(x); 100 else rotate(y); 101 } 102 rotate(x); 103 } 104 } 105 int top(int x){down(x);while(c[x][0])x=c[x][0],down(x);return x;} 106 void access(int x) 107 { 108 int t=0; 109 while(x) 110 { 111 splay(x); 112 if(c[x][1])addson(top(c[x][1]),1); 113 if(t)addson(top(t),-1); 114 c[x][1]=t;t=x;x=fa[x]; 115 } 116 } 117 void makeroot(int x){splay(x);rt=x;flip(x);} 118 }lct; 119 void dfs(int x) 120 { 121 in[x]=++dfn; 122 sgt.add(1,in[x],in[x],deep[x]); 123 for(int i=1;(1<<i)<=deep[x];i++) 124 fa[x][i]=fa[fa[x][i-1]][i-1]; 125 for(int i=first[x];i;i=e[i].next) 126 { 127 int to=e[i].to; 128 if(to==fa[x][0])continue; 129 fa[to][0]=lct.fa[to]=x; 130 deep[to]=deep[x]+1;dfs(to); 131 } 132 out[x]=dfn; 133 } 134 double request(int x) 135 { 136 if(x==rt)return 1.0*sgt.query(1,1,n)/n; 137 if(in[rt]>=in[x]&&out[rt]<=out[x]) 138 { 139 int t=find(x,rt);LL sum=0; 140 if(in[t]>1)sum+=sgt.query(1,1,in[t]-1); 141 if(out[t]<n)sum+=sgt.query(1,out[t]+1,n); 142 return 1.0*sum/(n-(out[t]-in[t]+1)); 143 } 144 return 1.0*sgt.query(1,in[x],out[x])/(out[x]-in[x]+1); 145 } 146 int main() 147 { 148 n=read();m=read(); 149 for(int i=1;i<n;i++)x=read(),y=read(),ins(x,y),ins(y,x); 150 sgt.build(1,1,n);deep[1]=1;dfs(1); 151 while(m--) 152 { 153 scanf("%s",op);x=read(); 154 if(op[2]==‘Q‘)printf("%.10lf\n",request(x)); 155 else 156 { 157 lct.access(x); 158 if(op[2]==‘C‘)lct.makeroot(x); 159 } 160 } 161 return 0; 162 }
〖相关资料〗
〖相关题目〗
1.【bzoj2286】[Sdoi2011]消耗战
题意:给定n个点的带边权树,每次询问给定ki个特殊点,求隔离点1和特殊点的最小代价。
分析:ONION_CYCの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=250005; 7 const int inf=0x3f3f3f3f; 8 int n,m,x,y,w,tim,cnt,cntv; 9 int a[N<<1],first[N],firstv[N],st[N]; 10 int dfn[N],out[N],deep[N],fa[N][20]; 11 LL val[N]; 12 bool f[N]; 13 struct edge{int to,next,w;}e[N<<1],ev[N<<1]; 14 int read() 15 { 16 int x=0,f=1;char c=getchar(); 17 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 18 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 19 return x*f; 20 } 21 void ins(int u,int v,int w){e[++cnt]=(edge){v,first[u],w};first[u]=cnt;} 22 void insv(int u,int v){ev[++cntv]=(edge){v,firstv[u],0};firstv[u]=cntv;} 23 void dfs(int x) 24 { 25 dfn[x]=++tim; 26 for(int i=1;(1<<i)<=deep[x];i++) 27 fa[x][i]=fa[fa[x][i-1]][i-1]; 28 for(int i=first[x];i;i=e[i].next) 29 { 30 int to=e[i].to; 31 if(to==fa[x][0])continue; 32 fa[to][0]=x; 33 deep[to]=deep[x]+1; 34 val[to]=min(val[x],1ll*e[i].w); 35 dfs(to); 36 } 37 out[x]=tim; 38 } 39 int lca(int x,int y) 40 { 41 if(deep[x]<deep[y])swap(x,y); 42 int d=deep[x]-deep[y]; 43 for(int i=0;(1<<i)<=d;i++) 44 if((1<<i)&d)x=fa[x][i]; 45 if(x==y)return x; 46 for(int i=18;i>=0;i--) 47 if((1<<i)<=deep[x]&&fa[x][i]!=fa[y][i]) 48 x=fa[x][i],y=fa[y][i]; 49 return fa[x][0]; 50 } 51 bool cmp(int a,int b){return dfn[a]<dfn[b];} 52 bool check(int x,int y){return dfn[x]<=out[y];} 53 LL dp(int x) 54 { 55 if(f[x])return val[x];LL sum=0; 56 for(int i=firstv[x];i;i=ev[i].next)sum+=dp(ev[i].to); 57 return min(val[x],sum); 58 } 59 void work() 60 { 61 int k=read(),tmp=k; 62 for(int i=1;i<=k;i++)a[i]=read(),f[a[i]]=true; 63 sort(a+1,a+k+1,cmp); 64 for(int i=1;i<tmp;i++)a[++k]=lca(a[i],a[i+1]); 65 sort(a+1,a+k+1,cmp); 66 k=unique(a+1,a+k+1)-a-1; 67 for(int i=1;i<=k;i++)firstv[a[i]]=0; 68 cntv=0;int top=0; 69 for(int i=1;i<=k;i++) 70 { 71 while(top&&!check(a[i],st[top]))top--; 72 if(top)insv(st[top],a[i]); 73 st[++top]=a[i]; 74 } 75 printf("%lld\n",dp(a[1])); 76 for(int i=1;i<=k;i++)f[a[i]]=false; 77 } 78 int main() 79 { 80 n=read(); 81 for(int i=1;i<n;i++) 82 { 83 x=read();y=read();w=read(); 84 ins(x,y,w);ins(y,x,w); 85 } 86 val[1]=1e16;dfs(1); 87 m=read(); 88 while(m--)work(); 89 return 0; 90 }
2.【bzoj3572】[Hnoi2014]世界树
题意:给定n个点的树,m次询问,每次给定ki个特殊点,一个点会被最近的特殊点控制,询问每个特殊点控制多少点。
分析:hzwerの博客
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 using namespace std; 6 const int N=300005; 7 int n,m,x,y,cnt,tim,top; 8 int a[N],b[N],bel[N],first[N],st[N],rem[N],f[N]; 9 int size[N],dfn[N],out[N],deep[N],fa[N][20]; 10 struct edge{int to,next;}e[N<<1]; 11 int read() 12 { 13 int x=0,f=1;char c=getchar(); 14 while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} 15 while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} 16 return x*f; 17 } 18 void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;} 19 void dfs(int x) 20 { 21 dfn[x]=++tim;size[x]=1; 22 for(int i=1;(1<<i)<=deep[x];i++) 23 fa[x][i]=fa[fa[x][i-1]][i-1]; 24 for(int i=first[x];i;i=e[i].next) 25 { 26 int to=e[i].to; 27 if(to==fa[x][0])continue; 28 deep[to]=deep[x]+1; 29 fa[to][0]=x;dfs(to); 30 size[x]+=size[to]; 31 } 32 out[x]=tim; 33 } 34 int lca(int x,int y) 35 { 36 if(deep[x]<deep[y])swap(x,y); 37 int d=deep[x]-deep[y]; 38 for(int i=0;(1<<i)<=d;i++) 39 if((1<<i)&d)x=fa[x][i]; 40 if(x==y)return x; 41 for(int i=18;i>=0;i--) 42 if((1<<i)<=deep[x]&&fa[x][i]!=fa[y][i]) 43 x=fa[x][i],y=fa[y][i]; 44 return fa[x][0]; 45 } 46 bool cmp(int a,int b){return dfn[a]<dfn[b];} 47 int dis(int a,int b){return deep[a]+deep[b]-2*deep[lca(a,b)];} 48 bool check(int x,int y){return dfn[x]<=out[y];} 49 void dfs1(int x) 50 { 51 rem[x]=size[x]; 52 for(int i=first[x];i;i=e[i].next) 53 { 54 int to=e[i].to;dfs1(to); 55 if(!bel[to])continue; 56 int d1=dis(x,bel[to]),d2=dis(x,bel[x]); 57 if(!bel[x]||d1<d2||(d1==d2&&bel[to]<bel[x]))bel[x]=bel[to]; 58 } 59 } 60 void dfs2(int x) 61 { 62 for(int i=first[x];i;i=e[i].next) 63 { 64 int to=e[i].to; 65 int d1=dis(to,bel[x]),d2=dis(to,bel[to]); 66 if(!bel[to]||d1<d2||(d1==d2&&bel[x]<bel[to]))bel[to]=bel[x]; 67 dfs2(to); 68 } 69 } 70 void solve(int a,int b) 71 { 72 int x=b,mid=b; 73 for(int i=18;i>=0;i--) 74 if(deep[fa[x][i]]>deep[a])x=fa[x][i]; 75 rem[a]-=size[x]; 76 if(bel[a]==bel[b]) 77 { 78 f[bel[a]]+=size[x]-size[b]; 79 return; 80 } 81 for(int i=18;i>=0;i--) 82 { 83 int y=fa[mid][i]; 84 if(deep[y]<=deep[a])continue; 85 int d1=dis(bel[a],y),d2=dis(bel[b],y); 86 if(d2<d1||(d1==d2&&bel[b]<bel[a]))mid=y; 87 } 88 f[bel[a]]+=size[x]-size[mid]; 89 f[bel[b]]+=size[mid]-size[b]; 90 } 91 void work() 92 { 93 int k=read(),tmp=k; 94 for(int i=1;i<=k;i++)a[i]=b[i]=read(); 95 for(int i=1;i<=k;i++)bel[a[i]]=a[i]; 96 sort(a+1,a+k+1,cmp); 97 for(int i=1;i<tmp;i++)a[++k]=lca(a[i],a[i+1]); 98 a[++k]=1;sort(a+1,a+k+1,cmp); 99 k=unique(a+1,a+k+1)-a-1; 100 cnt=top=0; 101 for(int i=1;i<=k;i++)first[a[i]]=0; 102 for(int i=1;i<=k;i++) 103 { 104 while(top&&!check(a[i],st[top]))top--; 105 if(top)ins(st[top],a[i]); 106 st[++top]=a[i]; 107 } 108 dfs1(1);dfs2(1); 109 for(int i=1;i<=k;i++) 110 for(int j=first[a[i]];j;j=e[j].next) 111 solve(a[i],e[j].to); 112 for(int i=1;i<=k;i++)f[bel[a[i]]]+=rem[a[i]]; 113 for(int i=1;i<=tmp;i++)printf("%d ",f[b[i]]); 114 printf("\n"); 115 for(int i=1;i<=k;i++)f[a[i]]=bel[a[i]]=0; 116 } 117 int main() 118 { 119 n=read(); 120 for(int i=1;i<n;i++) 121 { 122 x=read();y=read(); 123 ins(x,y);ins(y,x); 124 } 125 dfs(1);m=read(); 126 while(m--)work(); 127 return 0; 128 }
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原文地址:https://www.cnblogs.com/zsnuo/p/8308638.html