题目大意:有一张地图,地图上面有黄金,现在要求你用最少的1*2或者2*1的布条将这些黄金覆盖
解题思路:黄金分两个集合,能共用同一条布条的黄金连边,接着求出最大匹配数,最大匹配数/2就是共用一块布条的黄金了,所以最后的答案就是黄金数量 -最大匹配数
#include <cstdio>
#include <cstring>
const int N = 25;
int gold[N * N], g[N*N][N*N], left[N * N];
char map[N][N];
int n, m, cnt, cas = 1;
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
bool vis[N * N];
void init() {
scanf("%d%d", &n, &m);
memset(gold, 0, sizeof(gold));
cnt = 0;
for (int i = 0; i < n; i++) {
scanf("%s", map[i]);
for (int j = 0; j < m; j++)
if (map[i][j] == '*')
gold[i * 20 + j] = ++cnt;
}
memset(g, 0, sizeof(g));
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
if (map[i][j] == '*') {
for (int k = 0; k < 4; k++) {
int x = i + dir[k][0];
int y = j + dir[k][1];
if (x < 0 || y < 0 || x >= n || y >= m || map[x][y] == 'o') continue;
g[gold[i * 20 + j]][gold[x * 20 + y]] = true;
g[gold[x * 20 + y]][gold[i * 20 + j]] = true;
}
}
}
}
bool dfs(int u) {
for (int i = 1; i <= cnt; i++) {
if (g[u][i] && !vis[i]) {
vis[i] = true;
if (!left[i] || dfs(left[i])) {
left[i] = u;
return true;
}
}
}
return false;
}
void solve() {
memset(left, 0, sizeof(left));
int ans = 0;
for (int i = 1; i <= cnt; i++) {
memset(vis, 0, sizeof(vis));
if (dfs(i)) ans++;
}
// printf("cnt is %d ans is %d\n", cnt, ans);
printf("Case %d: %d\n", cas++, cnt - ans / 2);
}
int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}