链接:点击打开链接
题意:判断一个图是不是二分图,是的话求二分匹配最大匹配
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; int par[50005],ran[50005],x[50005],y[50005]; int found(int x){ if(par[x]!=x) par[x]=found(par[x]); return par[x]; } void unite(int x,int y){ x=found(x); y=found(y); if(x==y) return; if(ran[x]<ran[y]) par[x]=y; else{ par[y]=x; if(ran[x]==ran[y]) ran[x]++; } } bool same(int x,int y){ return found(x)==found(y); } int judge(int n,int m){ int i; unite(x[0],y[0]+n); unite(y[0],x[0]+n); for(i=1;i<m;i++){ if(same(x[i],y[i])||same(x[i]+n,y[i]+n)) return 0; unite(x[i],y[i]+n); unite(y[i],x[i]+n); } return 1; } //并查集判断是否是二分图,跟并查集食物链 struct node{ //那道题很像,可以看下面的链接 int u,v,cap; //http://blog.csdn.net/stay_accept/article/details/48490073 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; } //网络流模板 int main(){ int i,j,n,m; while(scanf("%d%d",&n,&m)!=EOF){ for(i=0;i<3*n;i++) par[i]=i,ran[i]=0; for(i=0;i<m;i++) scanf("%d%d",&x[i],&y[i]); //将边存起来用并查集判断是否可以构成二分图 if(!judge(n,m)){ puts("No"); continue; } 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); for(i=0;i<m;i++){ //建双向边避免一个点呗同时使用 addedge(x[i],y[i]+n,1); addedge(y[i],x[i]+n,1); } printf("%d\n",dinic()/2); //因为是双向边所以除以2 } return 0; }