Bzoj 2150(DAG的最小不相交路径覆盖)

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Bzoj 2150
用二分图最大匹配求DAG的最小不相交路径覆盖，答案为 原图的节点数 $-$ 新图的二分图最大匹配。算是模板题，注意高山不能算点，要注意特判，具体解法看这里

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define Uint unsigned int
#define db double
using namespace std;
const int MAXN = 50 + 5;
int dx[4], dy[4];
int mp[MAXN][MAXN], num[MAXN][MAXN], m, n, R, C;
char s[MAXN];
int cnt, newn, lk[MAXN * MAXN], vis[MAXN * MAXN];
vector<int> G[MAXN * MAXN];
bool hungary(int u) {
    for (Uint i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (vis[v] != cnt) {
            vis[v] = cnt;
            if (!lk[v] || hungary(lk[v])) {
                lk[v] = u;
                return true;
            }
        }
    } 
    return false;
}
void clean() {
    newn = 0, ms(mp, 0), ms(num, 0);
    for (int i = 0; i <= n * n; i++) G[i].clear(), lk[i] = vis[i] = 0;
}
void solve() {
    clean();
    for (int i = 1; i <= m; i++) {
        scanf("%s", s + 1);
        for (int j = 1; j <= n; j++) {
            if (s[j] == 'x') mp[i][j] = 1; else num[i][j] = ++newn; 
            //坐标转顶点编号新技巧，这里不能再用坐标转编号的数学方法了，因为是高山的话这里在新图中不算是一个点 
        }
    }
    dx[0] = R, dx[1] = R, dx[2] = C, dx[3] = C;
    dy[0] = -C, dy[1] = C, dy[2] = -R, dy[3] = R;//偏移坐标 
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (!mp[i][j]) 
            for (int qaq = 0; qaq < 4; qaq++) {
                int tx = dx[qaq] + i, ty = dy[qaq] + j;
                if (tx > 0 && ty > 0 && tx <= m && ty <= n && !mp[tx][ty]) G[num[i][j]].push_back(num[tx][ty]);
            }
        }
    }
    int ans = 0;
    for (int i = 1; i <= newn; i++) {
        if (hungary(cnt = i)) ans++;
    }
    printf("%d\n", newn - ans);
}
int main() {
    #ifndef ONLINE_JUDGE 
    freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
    #endif
    scanf("%d%d%d%d", &m, &n, &R, &C), solve();
    return 0;
}
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