标签：exp alt for its art view work while 线段
（压根就没做几道题，无地自容QAQ）
gym101138 J
直接上树链剖分+线段树，注意处理好端点问题（其实有（Q+N）logN的做法）
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 struct way{int po,next;} e[200010]; 6 struct node{ll lm,rm,s,mx;} tr[100010*4]; 7 int s[100010],a[100010],b[100010],c[100010],be[100010],d[100010],p[100010],fa[100010]; 8 int n,m,len,t,q; 9 const ll mi=-1000000000ll*100001ll; 10 const node inf={mi,mi,0,mi}; 11 void add(int x, int y) 12 { 13 e[++len].po=y; 14 e[len].next=p[x]; 15 p[x]=len; 16 } 17 18 void dfs1(int x) 19 { 20 s[x]=1; 21 for (int i=p[x]; i; i=e[i].next) 22 { 23 int y=e[i].po; 24 if (fa[x]!=y) 25 { 26 fa[y]=x; d[y]=d[x]+1; 27 dfs1(y); 28 s[x]+=s[y]; 29 } 30 } 31 } 32 33 void dfs2(int x) 34 { 35 c[x]=++t; 36 b[t]=x; 37 int q=0; 38 for (int i=p[x]; i; i=e[i].next) 39 { 40 int y=e[i].po; 41 if (c[y]==0&&s[q]<s[y]) q=y; 42 } 43 if (q) 44 { 45 be[q]=be[x]; 46 dfs2(q); 47 } 48 for (int i=p[x]; i; i=e[i].next) 49 { 50 int y=e[i].po; 51 if (fa[x]!=y&&y!=q) 52 { 53 be[y]=y; 54 dfs2(y); 55 } 56 } 57 } 58 59 void update(node &a,node b,node c) 60 { 61 a.s=b.s+c.s; 62 a.lm=max(b.lm,b.s+c.lm); 63 a.rm=max(c.rm,c.s+b.rm); 64 a.mx=max(b.rm+c.lm,max(b.mx,c.mx)); 65 a.mx=max(a.mx,max(a.rm,a.lm)); 66 } 67 68 void build(int i,int l,int r) 69 { 70 if (l==r) tr[i].mx=tr[i].s=tr[i].lm=tr[i].rm=a[b[l]]; 71 else { 72 int m=(l+r)>>1; 73 build(i*2,l,m); 74 build(i*2+1,m+1,r); 75 update(tr[i],tr[i*2],tr[i*2+1]); 76 } 77 } 78 79 node ask(int i,int l,int r,int x,int y) 80 { 81 if (x<=l&&y>=r) return tr[i]; 82 else { 83 int m=(l+r)>>1; 84 if (x<=m&&y>m) 85 { 86 node s,a=ask(i*2,l,m,x,y),b=ask(i*2+1,m+1,r,x,y); 87 update(s,a,b); 88 return s; 89 } 90 else if (x<=m) return ask(i*2,l,m,x,y); 91 else if (y>m) return ask(i*2+1,m+1,r,x,y); 92 } 93 } 94 95 ll work(int x,int y) 96 { 97 node sx=inf,sy=inf,s; 98 while (be[x]!=be[y]) 99 { 100 if (d[be[x]]>=d[be[y]]) 101 { 102 update(sx,ask(1,1,n,c[be[x]],c[x]),sx); 103 x=fa[be[x]]; 104 } 105 else { 106 update(sy,ask(1,1,n,c[be[y]],c[y]),sy); 107 y=fa[be[y]]; 108 } 109 } 110 swap(sx.lm,sx.rm); 111 if (c[x]>c[y]) 112 { 113 s=ask(1,1,n,c[y],c[x]); swap(s.lm,s.rm); 114 update(sx,sx,s); 115 } 116 else update(sy,ask(1,1,n,c[x],c[y]),sy); 117 update(s,sx,sy); 118 return s.mx; 119 } 120 121 int main() 122 { 123 scanf("%d",&n); 124 int x,y; 125 for (int i=1; i<n; i++) 126 { 127 scanf("%d%d",&x,&y); 128 add(x,y); 129 add(y,x); 130 } 131 for (int i=1; i<=n; i++) scanf("%d",&a[i]); 132 dfs1(1); 133 t=0; be[1]=1; 134 dfs2(1); 135 build(1,1,n); 136 scanf("%d",&q); 137 for (int i=1; i<=q; i++) 138 { 139 scanf("%d%d",&x,&y); 140 printf("%I64d\n",work(x,y)); 141 } 142 }
cf739 c
先差分，把差为0处理掉，就相当于拼接单调不下降序列，线段树维护之
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 struct node{int lm,rm,mx;} tr[300010*4]; 6 int a[300010],n,q; 7 ll b[300010]; 8 9 int dcmp(ll a) 10 { 11 if (a>0) return 1; 12 if (a<0) return -1; 13 return 0; 14 } 15 16 void update(int i,int l,int r) 17 { 18 int m=(l+r)>>1; 19 tr[i].mx=max(tr[i*2].mx,tr[i*2+1].mx); 20 if (dcmp(b[m])<=dcmp(b[m+1])) tr[i].mx=max(tr[i].mx,tr[i*2].rm+tr[i*2+1].lm); 21 tr[i].lm=tr[i*2].lm; 22 if (tr[i].lm==m-l+1&&dcmp(b[m])<=dcmp(b[m+1])) tr[i].lm=tr[i].lm+tr[i*2+1].lm; 23 tr[i].rm=tr[i*2+1].rm; 24 if (tr[i].rm==r-m&&dcmp(b[m])<=dcmp(b[m+1])) tr[i].rm=tr[i].rm+tr[i*2].rm; 25 } 26 27 void build(int i,int l,int r) 28 { 29 if (l==r) 30 { 31 int w=dcmp(b[l])!=0; 32 tr[i]=(node){w,w,w}; 33 } 34 else { 35 int m=(l+r)>>1; 36 build(i*2,l,m); 37 build(i*2+1,m+1,r); 38 update(i,l,r); 39 } 40 } 41 42 void work(int i,int l,int r,int x,int z) 43 { 44 if (l==r) 45 { 46 b[l]+=z; 47 int w=dcmp(b[l])!=0; 48 tr[i]=(node){w,w,w}; 49 } 50 else { 51 int m=(l+r)>>1; 52 if (x<=m) work(i*2,l,m,x,z); else work(i*2+1,m+1,r,x,z); 53 update(i,l,r); 54 } 55 } 56 57 int main() 58 { 59 scanf("%d",&n); 60 for (int i=1; i<=n; i++) scanf("%d",&a[i]); 61 for (int i=1; i<n; i++) b[i]=a[i]-a[i+1]; 62 if (n>1) build(1,1,n-1); 63 scanf("%d",&q); 64 for (int i=1; i<=q; i++) 65 { 66 int l,r,z; 67 scanf("%d%d%d",&l,&r,&z); 68 if (n==1) puts("1"); 69 else { 70 if (l>1) work(1,1,n-1,l-1,-z); 71 if (r<n) work(1,1,n-1,r,z); 72 printf("%d\n",tr[1].mx+1); 73 } 74 75 } 76 }
gym101138h
表示1~b意味着选出的面额从小到大排列后，任意s(i-1)+1>=di，然后可以用meet in middle加速（细节颇多）
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 struct node{ 6 ll d,a;int id; 7 friend bool operator <(node a,node b) 8 { 9 return a.d<b.d; 10 } 11 } a[50],c[(1<<21)+5]; 12 int t,n1,n2,n,s1,s2,b,mi[(1<<21)+5],w[50]; 13 ll ans; 14 const ll inf=9223372036854775807ll; 15 16 void pdf(int i,int st, ll sd, ll sa) 17 { 18 if (i==n1+1) 19 { 20 c[++t]=(node){sd,sa,st}; 21 if (sd>=b&&ans>sa) 22 { 23 ans=sa; 24 s1=st; s2=0; 25 } 26 return; 27 } 28 pdf(i+1,st,sd,sa); 29 if (sd+1>=a[i].d) pdf(i+1,st|(1<<(i-1)),sd+a[i].d,sa+a[i].a); 30 } 31 32 void dfs(int i,int st,ll sd,ll sa,ll key) 33 { 34 if (i==n2+1) 35 { 36 node x=(node){max(key,b-sd),0,0}; 37 int pos=lower_bound(c+1,c+1+t,x)-c; 38 if (pos==t+1) return; 39 if (c[mi[pos]].a+sa<ans) 40 { 41 s1=c[mi[pos]].id; s2=st; 42 ans=c[mi[pos]].a+sa; 43 } 44 return; 45 } 46 dfs(i+1,st,sd,sa,key); 47 dfs(i+1,st|(1<<(i-1)),sd+a[n1+i].d,sa+a[n1+i].a,max(key,a[i+n1].d-sd-1)); 48 } 49 50 int main() 51 { 52 scanf("%d%d",&n,&b); 53 for (int i=1; i<=n; i++) 54 { 55 int x,y; 56 scanf("%d%d",&x,&y); 57 a[i]=(node){x,y,i}; 58 } 59 sort(a+1,a+1+n); 60 ans=inf,s1=0,s2=0; 61 n1=n/2+(n%2),n2=n-n1,t=0; 62 pdf(1,0,0,0); 63 sort(c+1,c+1+t); 64 mi[t]=t; 65 for (int i=t-1; i; i--) 66 mi[i]=c[mi[i+1]].a>=c[i].a?i:mi[i+1]; 67 dfs(1,0,0,0,0); 68 if (ans<inf) 69 { 70 int s=0; 71 for (int i=0; i<n1; i++) 72 if (s1&(1<<i)) w[++s]=a[i+1].id; 73 for (int i=0; i<n2; i++) 74 if (s2&(1<<i)) w[++s]=a[i+n1+1].id; 75 printf("%d\n",s); 76 for (int i=1; i<=s; i++) 77 { 78 printf("%d",w[i]); 79 if (i<s) printf(" "); else puts(""); 80 } 81 } 82 else puts("-1"); 83 }
之前还有些题有意思的以后再放吧
标签：exp alt for its art view work while 线段
原文地址：http://www.cnblogs.com/phile/p/6123919.html
