题目链接：https://vjudge.net/problem/HDU-5093
按照行和列分别标注*的id，合并同行或同列相邻的块，二分图两部分分别是行和列，某一点(i,j)则连一条rid(i,j)到cid(i,j)的边。跑最大匹配。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 typedef struct pair<int, int> pii; 5 const int maxn = 110; 6 const int maxm = 3030; 7 const int inf = 0x3f3f3f3f; 8 9 int n, m; 10 char mp[maxn][maxn]; 11 12 int nu, nv, dist; 13 int Mx[maxm], My[maxm], dx[maxm], dy[maxm], vis[maxm], G[maxm][maxm]; 14 15 bool Dfs(int u){ 16 for(int v = 1; v <= nv; v++) 17 if(!vis[v] && G[u][v] && dy[v] == dx[u] + 1){ 18 vis[v] = 1; 19 if(My[v] != -1 && dy[v] == dist) 20 continue; 21 if(My[v] == -1|| Dfs(My[v])){ 22 My[v] = u; 23 Mx[u] = v; 24 return 1; 25 } 26 } 27 return 0; 28 } 29 bool Search(){ 30 queue<int> Q; 31 dist = inf; 32 memset(dx, -1, sizeof(dx)); 33 memset(dy, -1, sizeof(dy)); 34 for(int i = 1; i <= nu; i++) 35 if(Mx[i] == -1){ 36 Q.push(i); 37 dx[i] = 0; 38 } 39 while(!Q.empty()){ 40 int u = Q.front(); 41 Q.pop(); 42 if(dx[u] > dist) 43 break; 44 for(int v = 1; v <= nv; v++) 45 if(G[u][v] && dy[v] == -1){ 46 dy[v] = dx[u] + 1; 47 if(My[v] == -1) 48 dist = dy[v]; 49 else{ 50 dx[My[v]] = dy[v] + 1; 51 Q.push(My[v]); 52 } 53 } 54 } 55 return dist != inf; 56 } 57 int MaxMatch() { 58 int res = 0; 59 memset(Mx, -1, sizeof(Mx)); 60 memset(My, -1, sizeof(My)); 61 while(Search()){ 62 memset(vis, 0, sizeof(vis)); 63 for(int i = 0; i < nu; i++) 64 if(Mx[i] == -1&& Dfs(i)) 65 res++; 66 } 67 return res; 68 } 69 70 int rid[maxn][maxn], cid[maxn][maxn]; 71 72 73 int main() { 74 // freopen("in", "r", stdin); 75 int T; 76 scanf("%d", &T); 77 while(T--) { 78 scanf("%d%d",&n,&m); 79 for(int i = 0; i < n; i++) { 80 scanf("%s", mp[i]); 81 } 82 memset(G, 0, sizeof(G)); 83 memset(rid, 0, sizeof(rid)); 84 memset(cid, 0, sizeof(cid)); 85 int cnt = 1; 86 for(int i = 0; i < n; i++) { 87 for(int j = 0; j < m; j++) { 88 if(mp[i][j] == '*') rid[i][j] = cnt; 89 if(mp[i][j] == '#') rid[i][j] = cnt++; 90 } 91 cnt++; 92 } 93 nu = cnt; 94 cnt = 1; 95 for(int j = 0; j < m; j++) { 96 for(int i = 0; i < n; i++) { 97 if(mp[i][j] == '*') cid[i][j] = cnt; 98 if(mp[i][j] == '#') cid[i][j] = cnt++; 99 } 100 cnt++; 101 } 102 nv = cnt; 103 for(int i = 0; i < n; i++) { 104 for(int j = 0; j < m; j++) { 105 if(mp[i][j] == '*') { 106 G[rid[i][j]][cid[i][j]] = 1; 107 } 108 } 109 } 110 printf("%d ", MaxMatch()); 111 } 112 return 0; 113 }