Hdu1151 air raid-minimum path coverage

For a dag, a person can take a path, or at a point (the path length is 0), ask how many people need to cover all the points.

Based on the nature of the bipartite graph:

Minimum path coverage of Dag. After each vertex is split, find the maximum number of matching m and the result is n-M. When finding a specific path, follow the matching edge, match edge I → j ', J → K', K → l '.... form a directed path.

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include<queue>const int maxn=125;using namespace std;int mx[maxn],my[maxn],vis[maxn],e[maxn][maxn],n,m;int path(int u){    int i;    for(i=1;i<=n;i++)    {        if(e[u][i]&&!vis[i])        {            vis[i]=1;            if(my[i]==-1||path(my[i]))            {                mx[u]=i;                my[i]=u;                return 1;            }        }    }    return 0;}int hungry(){    int res=0;    memset(mx,-1,sizeof mx);    memset(my,-1,sizeof my);    for(int i=1;i<=n;i++)    {        if(mx[i]==-1)        {            memset(vis,0,sizeof vis);            res+=path(i);        }    }    return res;}int main(){    int ans,T,a,b,i;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        memset(e,0,sizeof e);        while(m--)        {            scanf("%d%d",&a,&b);            e[a][b]=1;        }        ans=hungry();        printf("%d\n",n-ans);    }    return 0;}