poj2226 二分图经典建图求最小点覆盖

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题意：：给出一个N行M列的图，’*’代表稀泥，’.’代表草地，现在要用一些木板把所有的稀泥盖住，但是不能盖住草地。一张木板只能盖住一行或者一列中的一部分，求至少要用多少木板把所有的稀泥盖住。

把横向连续的点压在一起作为L，把列向连续的点压在一起作为R，如果两者有交集，就从L到R连一条边。求最小点覆盖(==最大匹配数)

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
const int N = 55;
const int M = N*N;
char a[N][N];
int row[N][N],col[N][N];
int R,C;
bool vis[M];
int linker[M];
int to[M],nxt[M];
int head[M],tot;
int n, m;

void addedge(int u, int v)
{
    ++tot;
    to[tot] = v;
    nxt[tot] = head[u];
    head[u] = tot;
}

bool dfs(int u)
{
    for(int i = head[u]; ~i; i = nxt[i])
    {
        int v = to[i];
        if(!vis[v])
        {
            vis[v] = 1;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v] = u;
                return 1;
            }
        }
    }
    return 0;
}

int hungary()
{
    int ret = 0;
    memset(linker, -1, sizeof(linker));
    for(int i = 1; i <= R; ++i)
    {
        memset(vis, 0, sizeof(vis));
        if(dfs(i))
            ++ret;
    }
    return ret;
}

int main()
{
    while(~scanf("%d %d", &n, &m))
    {
        memset(row, 0, sizeof(row));
        memset(col, 0, sizeof(col));
        memset(head, -1, sizeof(head));
        tot = -1;
        R = C = 0;
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
            {
                cin >> a[i][j];
                if(a[i][j]=='*')
                {
                    if(row[i][j-1]) row[i][j] = row[i][j-1];
                    else row[i][j] = ++R;
                    if(col[i-1][j]) col[i][j] = col[i-1][j];
                    else col[i][j] = ++C;
                }
            }
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= m; ++j)
                if(a[i][j]=='*')
                    addedge(row[i][j], col[i][j]);
        printf("%d\n", hungary());
    }
    return 0;
}