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呵呵!原来就是二分图的匹配问题好简单!#include <stdio.h>
#include <string.h>
#define size 150
int g[size][size];
int used[size];
int match[size];
int n,m;
int dfs(int x)
{
int i;
for(i=1;i<=n;i++)
{
if(!used[i]&&g[x][i])
{
used[i]=1;
if(match[i]==-1||dfs(match[i]))
{
match[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int cases,i,a,b,res;
scanf("%d",&cases);
while(cases--)
{
scanf("%d",&n);
scanf("%d",&m);
memset(g,0,sizeof(g));
memset(match,-1,sizeof(match));
for(i=1;i<=m;i++)
{
scanf("%d %d",&a,&b);
g[a][b]=1;
}
res=0;
for(i=1;i<=n;i++)
{
memset(used,0,sizeof(used));
if(dfs(i))
res++;
}
printf("%d\n",n-res);
}
return 1;
}
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