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#include <bits/stdc++.h>
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using namespace std;
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const int N = 510, M = N * N * 2;
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int dx[] = {0, 0, -1, 1};
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int dy[] = {-1, 1, 0, 0};
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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// 匈牙利算法
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int st[N], match[N];
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int dfs(int u) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (st[v]) continue;
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st[v] = 1;
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if (match[v] == -1 || dfs(match[v])) {
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match[v] = u;
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return 1;
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}
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}
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return 0;
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}
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char g[50][50];
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int id[50][50];
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("LightOJ1152.in", "r", stdin);
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#endif
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int T, n, m, cas = 0;
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scanf("%d", &T);
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while (T--) {
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// 初始化链式前向星
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memset(h, -1, sizeof h);
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idx = 0;
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scanf("%d%d", &n, &m);
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// 按char[]数组读入原始地图
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for (int i = 1; i <= n; i++) scanf(" %s", g[i] + 1);
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// 给每个金子编号,相当于离散化
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memset(id, 0, sizeof id);
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int cnt = 0;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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if (g[i][j] == '*') id[i][j] = ++cnt;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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if (id[i][j]) {
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for (int k = 0; k < 4; k++) {
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int tx = i + dx[k], ty = j + dy[k];
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if (!id[tx][ty]) continue;
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if (tx < 1 || tx > n || ty < 1 || ty > m) continue;
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add(id[i][j], id[tx][ty]); // 对每个金子，和它四个方向上的金子连边，没有就不连
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}
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}
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int res = 0;
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memset(match, -1, sizeof match);
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for (int i = 1; i <= cnt; i++) {
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memset(st, 0, sizeof st);
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if (dfs(i)) res++;
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}
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/*
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*由于对每个金子拆点，求出的最大匹配是双倍的，除以2才是真正的最大匹配，不除以2则是匹配的点，
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*那么节点数 - 匹配的点 = 落单的个数。明显落单的金子每个需要一个骨牌，匹配的点两个需要一个骨牌
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*于是，答案= 节点 - 匹配点 + 匹配点 / 2
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*/
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printf("Case %d: %d\n", ++cas, cnt - res / 2);
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}
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return 0;
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} |
