试题链接：https://www.nowcoder.com/acm/contest/76/A
【思路】
每个‘#’的右边和下边如果也是‘#’说明这两个点构成通路，以此重构一幅图，然后找二分图的最大匹配。
【代码】
#include<bits/stdc++.h> using namespace std; char mp[55][55]; bool vis[2505]; vector<int>G[2505]; int mp1[55][55], match[2505], n; bool dfs(int u) { for(int i = 0; i < G[u].size(); i++) { int v = G[u][i]; if(!vis[v]) { vis[v] = 1; if(match[v] == -1 || dfs(match[v])) { match[v] = u; return 1; } } } return 0; } int main() { int t, cas = 0; cin>>t; while(t--) { int num = 0; scanf("%d", &n); for(int i = 0; i < n; i++) scanf("%s", mp[i]); for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) if(mp[i][j] == ‘#‘) mp1[i][j] = ++num; for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) { if(mp[i][j] == ‘#‘) { if(mp[i+1][j] == ‘#‘) G[mp1[i][j]].push_back(mp1[i+1][j]), G[mp1[i+1][j]].push_back(mp1[i][j]); if(mp[i][j+1] == ‘#‘) G[mp1[i][j]].push_back(mp1[i][j+1]), G[mp1[i][j+1]].push_back(mp1[i][j]); } } memset(match, -1, sizeof match); int ans = 0; for(int i = 1; i <= num; i++) { memset(vis, 0, sizeof vis); if(dfs(i)) ans++; } printf("Case %d: %d\n", ++cas, ans/2); } return 0; }