大意：有n個男孩n個女孩，告訴你每個女孩喜歡哪些男孩，又告訴你女孩之間的存在一些朋友關系

一個女孩可以和她喜歡的男孩結婚也可以和她朋友喜歡的男孩結婚， 并且朋友關系可以傳遞

Once every girl finds their boyfriends they will start a new round of this game―marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.?
Now, here is the question for you, how many rounds can these 2n kids totally play this game?

問最后能進行多少輪game

分析：

用并查集處理處朋友的關系集合 ? 把每個人到喜歡的男孩和其朋友喜歡的男孩建邊

當最大匹配 == n 的時候 rounds++ 然后刪除匹配邊

求出round數即可

?

代碼：

1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6 7 const int maxn = 105; 8 const int INF = 1000000000; 9 10 int fa[maxn]; 11 void init(int x) { 12 for(int i = 0; i <= x; i++) { 13 fa[i] = i; 14 } 15 } 16 17 int find(int u) { 18 if(fa[u] == u) return u; 19 return fa[u] = find(fa[u]); 20 } 21 22 void unin(int u, int v) { 23 int fu = find(u); 24 int fv = find(v); 25 if(fu != fv) { 26 fa[fu] = fv; 27 } 28 } 29 30 31 int Link[maxn]; 32 int W[maxn][maxn]; 33 int vis[maxn]; 34 int n; 35 36 int Find(int i) { 37 for(int j = 1; j <= n; j++) { 38 if(W[i][j] && !vis[j]) { 39 vis[j] = 1; 40 if(Link[j] == -1 || Find(Link[j]) ) { 41 Link[j] = i; 42 return true; 43 } 44 } 45 } 46 return false; 47 } 48 49 bool Match() { 50 memset(Link, -1, sizeof(Link)); 51 for(int i = 1; i <= n; i++) { 52 memset(vis, 0, sizeof(vis)); 53 if(!Find(i)) return false; 54 } 55 return true; 56 } 57 58 int Solve() { 59 int cnt = 0; 60 while(1) { 61 if(Match() ) { 62 cnt ++; 63 for(int j = 1; j <= n; j++) { 64 W[Link[j]][j] = 0; 65 } 66 } else break; 67 } 68 return cnt; 69 } 70 71 vector<int> v[maxn * maxn]; 72 73 int main() { 74 int m, f; 75 int a, b; 76 int c, d; 77 int t; 78 scanf("%d", &t); 79 while(t--) { 80 scanf("%d %d %d", &n, &m, &f); 81 memset(W, 0, sizeof(W)); 82 init(n); 83 for(int i = 0; i <= n; i++) v[i].clear(); 84 for(int i = 1; i <= m; i++) { 85 scanf("%d %d", &a, &b); 86 v[a].push_back(b); 87 } 88 for(int i = 1; i <= f; i++) { 89 scanf("%d %d",&c, &d); 90 unin(c, d); 91 } 92 for(int i = 1; i <= n; i++) { 93 for(int j = 1; j <= n; j++) { 94 if(find(i) == find(j) ) { 95 for(int k = 0; k < v[j].size(); k++) { 96 W[i][v[j][k]] = 1; 97 } 98 } 99 } 100 } 101 printf("%d\n", Solve() ) ; 102 } 103 return 0; 104 } View Code

?

轉載于:https://www.cnblogs.com/zhanzhao/p/3962923.html

總結

以上是生活随笔為你收集整理的HDU 3081 Marriage Match II【并查集+二分图最大匹配】的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。