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题意:求n个点m条边的有向无环图的最小路径覆盖
代码:
#include <queue> #include <vector> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; const int INF=0x3f3f3f3f; struct node{ int u,v,cap; node(){} node(int u,int v,int cap):u(u),v(v),cap(cap){} }es[500005]; int R,S,T; int dis[50005],iter[50005]; vector<int> tab[50005]; void addedge(int u, int v, int cap){ tab[u].push_back(R); es[R++]=node(u,v,cap); tab[v].push_back(R); es[R++]=node(v,u,0); } int bfs(){ int i,h; queue<int> q; q.push(S); memset(dis,INF,sizeof(dis)); dis[S]=0; while(q.size()){ h=q.front(); q.pop(); for(i=0;i<tab[h].size();i++){ node &e=es[tab[h][i]]; if(e.cap>0&&dis[e.v]==INF){ dis[e.v]=dis[h]+1; q.push(e.v); } } } return dis[T]<INF; } int dfs(int x,int maxflow){ int flow; if(x==T) return maxflow; for(int &i=iter[x];i<tab[x].size();i++){ node &e=es[tab[x][i]]; if(dis[e.v]==dis[x]+1&&e.cap>0){ flow=dfs(e.v,min(maxflow,e.cap)); if(flow){ e.cap-=flow; es[tab[x][i]^1].cap+=flow; return flow; } } } return 0; } int dinic(){ int ans,flow; ans=0; while(bfs()){ memset(iter,0,sizeof(iter)); while(flow=dfs(S,INF)) ans+=flow; } return ans; } //网络流求二分图最大匹配 int main(){ int i,j,k,x,y,n,m; int s[505][505]; while(scanf("%d%d",&n,&m)!=EOF&&(n||m)){ R=0,S=0,T=2*n+1; for(i=S;i<=T;i++) tab[i].clear(); for(i=1;i<=n;i++) addedge(S,i,1); for(i=n+1;i<=2*n;i++) addedge(i,T,1); memset(s,0,sizeof(s)); for(i=0;i<m;i++){ scanf("%d%d",&x,&y); s[x][y]=1; //拆点但是是只能从左连向右,可以找出一些情况证明不能从左连向右 } for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) s[i][j]=s[i][j]+s[i][k]*s[k][j]; //求传递闭包避免某些情况,例如 for(i=1;i<=n;i++) //http://www.cnblogs.com/ka200812/archive/2011/07/31/2122641.html for(j=1;j<=n;j++) if(s[i][j]) addedge(i,j+n,1); printf("%d\n",n-dinic()); } return 0; }