hdu2444(二分图判断+二分图最大匹配)

#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;
}