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

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

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

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

 /*
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 */