p2319 [HNOI2006]超级英雄

传送门

分析

从1道m进行匹配，找到第一个不能继续匹配的点即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int n,m,g[1100][1100],used[1100],T,wh[1100],ans[1100],Ans;
inline bool work(int x){
    int i,j,k;
    for(i=1;i<=n;i++)
      if(g[x][i]&&used[i]!=T){
          used[i]=T;
          if(!wh[i]||work(wh[i])){
            wh[i]=x;
            ans[x]=i;
            return 1;
        }
      }
    return 0;
}
inline void go(){
    int i,j,k;
    for(i=1;i<=m;i++){
      ++T;
      if(work(i))Ans++;
        else return;
    }
}
int main(){
    int i,j,k;
    scanf("%d%d",&n,&m);
    for(i=1;i<=m;i++){
      int x,y;
      scanf("%d%d",&x,&y);
      x++,y++;
      g[i][x]=g[i][y]=1;
    }
    go();
    printf("%d\n",Ans);
    for(i=1;i<=Ans;i++)printf("%d\n",ans[i]-1);
    return 0;
}