[BZOJ4554][TJOI2016&HEOI2016]游戏（二分图匹配）

和[ZJOI2007]矩阵游戏是类似的思想。首先，在行上，把连续的不含#的连通块标记出来，如：
上图中用框标出的就是被标记出的连通块。
然后对列上也一样做。然后对于任意一个可以放炸弹的格子，从它所在的行连通块标号向所在的列连通块标号连一条边。然后跑一遍二分图最大匹配，即为答案。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}inline char get() {    char c; while ((c = getchar()) != '*' && c != 'x' && c != '#');    return c;}const int N = 55, M = 2505;int n, m, tot, ecnt, nxt[M], adj[M], go[M], my[M], row[N][N], col[N][N];bool vis[M]; char a[N][N];void add_edge(int u, int v) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v;}bool dfs(int u) {    for (int e = adj[u], v; e; e = nxt[e])        if (!vis[v = go[e]]) {            vis[v] = 1;            if (!my[v] || dfs(my[v])) {                my[v] = u;                return 1;            }        }    return 0;}int solve() {    int i, ans = 0;    for (i = 1; i <= tot; i++) {        memset(vis, 0, sizeof(vis));        if (dfs(i)) ans++;    }    return ans;}int main() {    int i, j; n = read(); m = read();    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++)        a[i][j] = get();    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) {        if (a[i][j] == '#') continue;        if (j == 1 || a[i][j - 1] == '#') tot++;        row[i][j] = tot;    }    for (j = 1; j <= m; j++) for (i = 1; i <= n; i++) {        if (a[i][j] == '#') continue;        if (i == 1 || a[i - 1][j] == '#') tot++;        col[i][j] = tot;    }    for (i = 1; i <= n; i++) for (j = 1; j <= m; j++) {        if (a[i][j] != '*') continue;        add_edge(row[i][j], col[i][j]);    }    printf("%d\n", solve());    return 0;}