4554: [Tjoi2016&Heoi2016]游戏 二分图匹配

题目链接：

http://www.lydsy.com/JudgeOnline/problem.php?id=4554

题解：

如果没有硬石头的话，就是’*‘点对应的行列建边，然后跑最大匹配

硬石头什么作用？它可以让同一行或同一列存在不只一个炸弹，因此我们可以将一个硬石头的上下拆成两列，左右拆成两行，然后就可以用经典的做法来跑最大匹配了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

const int maxn = 55;
const int maxs = 2555;
int n, m;

char str[maxn][maxn];
int _x[maxn][maxn], _y[maxn][maxn];
int _r, _c;
vector<int> G[maxs];

int lef[maxs], _t[maxs];
bool match(int u) {
    for (int j = 0; j < G[u].size(); j++) {
        int v = G[u][j];
        if (!_t[v]) {
            _t[v] = 1;
            if (lef[v]==-1 || match(lef[v])) {
                lef[v] = u;
                return true;
            }
        }
    }
    return false;
}

int BM() {
    memset(lef, -1, sizeof(lef));
    for (int i = 0; i < _r; i++) {
        memset(_t, 0, sizeof(_t));
        match(i);
    }
    int ret = 0;
    for (int i = 0; i < _c; i++) {
        if (lef[i] != -1) ret++;
    }
    return ret;
}

void init() {
    for (int i = 0; i < maxs; i++) G[i].clear();
}

int main() {
    while (scanf("%d%d", &n, &m) == 2 && n) {
        init();
        for (int i = 0; i < n; i++) scanf("%s", str[i]);
        _r = 0, _c = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (str[i][j] == '#') _r++;
                else if (str[i][j] == '*') _x[i][j] = _r;
            }
            _r++;
        }
        for (int j = 0; j < m; j++) {
            for (int i = 0; i < n; i++) {
                if (str[i][j] == '#') _c++;
                else if (str[i][j] == '*') _y[i][j] = _c;
            }
            _c++;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (str[i][j] == '*') {
                    G[_x[i][j]].push_back(_y[i][j]);
                }
            }
        }
        printf("%d\n", BM());
    }
    return 0;
}

/*
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*/