poj Antenna Placement（二分图的匹配）

是最小路径覆盖问题：最小路径覆盖=最小路径覆盖＝｜G｜－最大匹配数 
#include<stdio.h>
#include<string.h>
const int maxn = 40*10+10;
int maps[45][15],g[maxn][maxn],linker[maxn],vis[maxn],uN,vN;
int dfs(int u)
{
    for(int v = 1; v <= vN; v++)
        if(!vis[v] && g[u][v])
        {
            vis[v] = 1;
            if(linker[v] == -1 || dfs(linker[v]))
            {
                linker[v] = u;
                return 1;
            }
        }
    return 0;
}
int hungary()
{
    int sum = 0;
    memset(linker,-1,sizeof(linker));
    for(int u = 1; u <= uN; u++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(u))
            sum++;
    }
    return sum;
}
int main()
{
    int T,row,col;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%*c", &row,&col);
        char c;
        int num = 0;
        memset(maps,0,sizeof(maps));
        memset(g,0,sizeof(g));
        for(int i = 1; i <= row; i++)
        {
            for(int j = 1; j <= col; j++)
            {
                scanf("%c", &c);
                if(c == '*') maps[i][j] = ++num;
            }
            getchar();
        }
        uN = vN = num;
        for(int i = 1; i <= row; i++)
        {
            for(int j = 1; j <= col; j++)
            {
                if(maps[i][j])
                {
                    if(maps[i][j+1])
                        g[maps[i][j]][maps[i][j+1]] = 1;
                    if(maps[i][j-1])
                        g[maps[i][j]][maps[i][j-1]] = 1;
                    if(maps[i+1][j])
                        g[maps[i][j]][maps[i+1][j]] = 1;
                    if(maps[i-1][j])
                        g[maps[i][j]][maps[i-1][j]] = 1;
                }
            }
        }
        printf("%d\n", num-hungary()/2);
    }
    return 0;
}