标签：using 线段 lld date turn cst root name 二分图
#include <cstdio> #include <iostream> #include <cstring> using namespace std; int c[1000000],n,m; char s; void update(int p,int l,int r,int x,int add) { int m=(l+r) / 2; if (l==r) { c[p]+=add; return; } if (x<=m) update(p*2,l,m,x,add); if (x>m) update(p*2+1,m+1,r,x,add); c[p]=c[p<<1]+c[p<<1|1]; } int find(int p,int l,int r,int a,int b) { if (l==a && r==b) return c[p]; else { int m=(l+r) / 2; if (b<=m) return find(p*2,l,m,a,b); else if (a>m) return find(p*2+1,m+1,r,a,b); else return find(p*2,l,m,a,m)+find(p*2+1,m+1,r,m+1,b); } } int main() { scanf("%d",&n); scanf("%d",&m); for (int i=1;i<=m;i++) { cin>>s; if (s==‘M‘) { int x,y; scanf("%d%d",&x,&y); update(1,1,n,x,y); } if (s==‘C‘) { int x,y; scanf("%d%d",&x,&y); printf("%d\n",find(1,1,n,x,y)); } } }
线段树，并且单纯的线段树会超时，因为在将a到b的点全部加上c时，步骤太多，会超时。
需要优化。即lazy算法；
Lazy：
在将a~b点全部加c时，不要加到每个点，在表示区间的root结构体上增加一个inc域，将要加的值赋给这个inc域，然后就不要再往下了。
在求区间和时，将root中的inc值赋给要求的区间，并且将该节点的root置零。
#include <cstdio> #include <iostream> #include <cstring> #include <string> #define ll long long using namespace std; ll a[1000005],t[5000005],n,lazy[5000005],m; ll read() { ll s=0; char ch=getchar(); while (ch<‘0‘ || ch>‘9‘) ch=getchar(); while (ch>=‘0‘ && ch<=‘9‘) { s=(s << 1) + (s << 3) +ch-‘0‘;ch=getchar(); } return s; } void build(ll p,ll l,ll r) { if (l==r) { t[p]=a[l]; return; } ll m=(l+r)>>1; build(p<<1,l,m); build(p<<1|1,m+1,r); t[p]=t[p<<1]+t[p<<1|1]; } void pushdown(ll x,ll len) { if (lazy[x]!=0) { lazy[x<<1]+=lazy[x]; lazy[x<<1|1]+=lazy[x]; t[x<<1]+=(len-len/2)*lazy[x]; t[x<<1|1]+=(len/2)*lazy[x]; lazy[x]=0; } } void update(ll a,ll b,ll l,ll r,ll x,ll p) { if (a<=l && b>=r) { t[p]+=(r-l+1)*x; lazy[p]+=x; return; } if (lazy[p]!=0) pushdown(p,r-l+1); ll m=(l+r)/2; if (b<=m) update(a,b,l,m,x,p<<1); else if (a>m) update(a,b,m+1,r,x,p<<1|1); else update(a,m,l,m,x,p<<1),update(m+1,b,m+1,r,x,p<<1|1); t[p]=t[p<<1]+t[p<<1|1]; } ll find(ll a,ll b,ll l,ll r,ll p) { if (lazy[p]!=0) pushdown(p,r-l+1); if (a==l && b==r) return t[p]; ll m=(l+r)/2; if (b<=m) return find(a,b,l,m,p<<1); else if (a>m) return find(a,b,m+1,r,p<<1|1); else return find(a,m,l,m,p<<1)+find(m+1,b,m+1,r,p<<1|1); } int main() { n=read(); m=read(); for (int i=1;i<=n;i++) a[i]=read(); build(1,1,n); int x,y,z; for (int i=1;i<=m;i++) { x=read(); if (x==1) { x=read(),y=read(),z=read(); update(x,y,1,n,z,1); } else if (x==2) { y=read(),z=read(); printf("%lld\n",find(y,z,1,n,1)); } } }
#include <iostream> #include <cstring> #include <cstdio> #define inf 1000007 using namespace std; struct arr { int y, n; }f[1000005]; int ls[1000005], n, m, tot, e, ans, link[1000005]; bool b[1000005]; void add(int x, int y) { f[++tot].y = y; f[tot].n = ls[x]; ls[x] = tot; } bool find(int x) { for (int i = ls[x]; i; i = f[i].n) if (!b[f[i].y]) { b[f[i].y] = 1; if (!link[f[i].y] || find(link[f[i].y])) { link[f[i].y] = x; return true; } } return false; } void work() { for (int i = 1; i <= n; i++) { memset(b, 0, sizeof(b)); if (find(i)) ans++; } } int main() { int x, y; scanf("%d%d%d", &n, &m, &e); for (int i = 1; i <= e; i++) { scanf("%d%d", &x, &y); if (y > m) continue; add(x, y); } work(); printf("%d", ans); }
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; struct arr { int x,y,w; }f[200007]; int g[5007],tot,n,m; bool b[5007]; int cmp(arr a,arr b) { return a.w<b.w; } int find(int x) { if (g[x]==x) return x; g[x]=find(g[x]); return g[x]; } int main() { cin>>n>>m; int q,p,o,e=0; for (int i=1;i<=m;i++) { scanf("%d%d%d",&q,&p,&o); e++; f[e].x=q; f[e].y=p; f[e].w=o; } sort(f+1,f+e+1,cmp); for (int i=1;i<=n;i++) g[i]=i; int k=0; for (int i=1;i<=e;i++) { if (k==n-1) break; if (find(g[f[i].x])!=find(g[f[i].y])) { k++; tot+=f[i].w; g[find(f[i].x)]=find(f[i].y); } } if (k==n-1) printf("%d",tot); else printf("orz"); }
#include <cstdio> #include <cstring> #include <iostream> using namespace std; struct arr { int to,next,w; }f[5000007]; int n,m,s,d[1000007],ls[1000007],str[1000007]; bool b[1000007]; void spfa() { for (int i=1;i<=n;i++) d[i]=2147483647; d[s]=0; int head=0,tail=1; b[s]=1; str[1]=s; while (head<tail) { head++; int t=ls[str[head]]; while (t>0) { if (d[f[t].to]>d[str[head]]+f[t].w) { d[f[t].to]=d[str[head]]+f[t].w; if (!b[f[t].to]) { b[f[t].to]=1; tail++; str[tail]=f[t].to; } } t=f[t].next; } b[str[head]]=0; } } int main() { scanf("%d%d%d",&n,&m,&s); int p,q; for (int i=1;i<=m;i++) { scanf("%d%d%d",&q,&p,&f[i].w); f[i].to=p; f[i].next=ls[q]; ls[q]=i; } spfa(); for (int i=1;i<=n;i++) printf("%d ",d[i]); }
#include <cstdio> #include <cstring> #include <iostream> using namespace std; int n,m,f[10007]; int find(int x) { if (f[x]==x) return x; f[x]=find(f[x]); return f[x]; } int main() { scanf("%d%d",&n,&m); int q,p,z; for (int i=1;i<=n;i++) f[i]=i; for (int i=1;i<=m;i++) { scanf("%d%d%d",&z,&q,&p); if (z==1) { f[find(q)]=find(p); } if (z==2) { if (find(q)==find(p)) cout<<"Y"<<endl; else cout<<"N"<<endl; } } }
#include <cstdio> #include <cstring> #include <iostream> using namespace std; int a[1000005],b[1000005],p[1000005],n,m,next[1000005],ans; int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) { scanf("%d", &p[i]); a[i - 1] = p[i] - p[i - 1]; } for (int i = 1; i <= m; i++) { scanf("%d", &p[i]); b[i - 1] = p[i] - p[i - 1]; } int j = 0; for (int i = 2; i < m; i++) { while (j > 0 && b[j + 1] != b[i]) j = next[j]; if (b[j + 1] == b[i]) j += 1; next[i] = j; } j = 0; for (int i = 1; i < n; i++) { while (j > 0 && b[j + 1] != a[i]) j = next[j]; if (b[j + 1] == a[i]) j += 1; if (j == m - 1) { ans++; j = next[j]; } } printf("%d", ans); }
int pow4(int a,int b) { int r=1,base=a; while(b) { if(b&1) r*=base; base*=base; b>>=1; } return r; }
#include <cstdio> #include <iostream> using namespace std; struct arr { int x,y,next; }; arr f[3000]; int ls[40000],stack[40000],belong[40000],low[40000],dfn[40000],cnt,e,n,d,tot,rd[40000],cd[40000],ans,s1,s2; bool ins[40000]; void add(int x,int y) { e++; f[e].x=x; f[e].y=y; f[e].next=ls[x]; ls[x]=e; } void tarjan(int i) { int t=0; int j=0; d++; dfn[i]=d; low[i]=d; tot++; stack[tot]=i; t=ls[i]; ins[i]=1; while (t>0) { j=f[t].y; if (dfn[j]==0) { tarjan(j); if (low[i]>low[j]) low[i]=low[j]; } else if (ins[j]&&dfn[j]<low[i]) low[i]=dfn[j]; t=f[t].next; } if (dfn[i]==low[i]) { cnt++; do { j=stack[tot]; ins[j]=false; tot--; belong[j]=cnt; } while (i!=j); } } int main() { cin>>n; int j; e=0; for (int i=1;i<=n;i++) { while (scanf("%d",&j)&&j!=0) add(i,j); } for (int i=1;i<=n;i++) if (dfn[i]==0) tarjan(i); for (int i=1;i<=e;i++) if (belong[f[i].x]!=belong[f[i].y]) { cd[belong[f[i].x]]++; rd[belong[f[i].y]]++; } ans=0; s1=0; s2=0; for (int i=1;i<=cnt;i++) { if (rd[i]==0) {ans++; s1++;} if (cd[i]==0) s2++; } cout<<ans<<endl; if (cnt==1) {s1=0; s2=0;} if (s1>s2) cout<<s1<<endl; else cout<<s2<<endl; }
for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (i != j && i != k && j != k) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
#include <cstdio> #include <cstring> #include <iostream> using namespace std; struct arr {int x,y,n;}f[60007]; int ls[30007],n,m,e,deep[30007],fa[30007][30],ans; void add(int x, int y) { f[++e].x = x; f[e].y = y; f[e].n = ls[x]; ls[x] = e; } void dfs(int x, int la) { deep[x] = deep[la] + 1; fa[x][0] = la; for (int i = ls[x]; i ; i = f[i].n) { if (f[i].y != la) { dfs(f[i].y, x); } } } int lca(int q, int p) { if (deep[p] < deep[q]) swap(p,q); for (int j = 20; j >= 0; j--) if (deep[fa[p][j]] >= deep[q]) p = fa[p][j]; if (p == q) return p; for (int j = 20; j >= 0; j--) if (fa[p][j] != fa[q][j]) { p = fa[p][j], q = fa[q][j]; } return fa[p][0]; } int main() { scanf("%d", &n); int x, y; for (int i = 1; i <= n - 1; i++) { scanf("%d%d", &x, &y); add(x, y); add(y, x); } dfs(1, 0); for (int j = 1; j <= 20; j++) for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1]; scanf("%d", &m); scanf("%d", &x); m--; while (m--) { scanf("%d", &y); ans += deep[x] + deep[y] - deep[lca(x, y)] * 2; x = y; } printf("%d", ans); }
线段树、最短路径、最小生成树、并查集、二分图匹配、最近公共祖先--C++模板
标签：using 线段 lld date turn cst root name 二分图
原文地址：https://www.cnblogs.com/jiangxin/p/11840282.html
