HDU 3081 Marriage Match II

Marriage Match II

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 3081
64-bit integer IO format: %I64d      Java class name: Main

 

 

Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids. 
Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her boyfriend when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend. 
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?

 

Input

There are several test cases. First is a integer T, means the number of test cases. 
Each test case starts with three integer n, m and f in a line (3<=n<=100,0<m<n*n,0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other. 
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.

 

Output

For each case, output a number in one line. The maximal number of Marriage Match the children can play.

 

Sample Input

1

4 5 2

1 1

2 3

3 2

4 2

4 4

1 4

2 3

Sample Output

2

Source

HDU 2nd “Vegetable-Birds Cup” Programming Open Contest

 

解题：求最大匹配数的种数，就是求最大匹配的方案数，姿势？二分+最大流。。。二超级源点到女生的弧的容量，表示女生最多可以同几名男生。。。oo...

 

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <climits>

  7 #include <vector>

  8 #include <queue>

  9 #include <cstdlib>

 10 #include <string>

 11 #include <set>

 12 #include <stack>

 13 #define LL long long

 14 #define pii pair<int,int>

 15 #define INF 0x3f3f3f3f

 16 using namespace std;

 17 const int maxn = 300;

 18 struct arc {

 19     int to,flow,next;

 20     arc(int x = 0,int y = 0,int z = -1) {

 21         to = x;

 22         flow = y;

 23         next = z;

 24     }

 25 };

 26 arc e[100000];

 27 bool g[maxn][maxn];

 28 int head[maxn],d[maxn],cur[maxn];

 29 int S,T,tot,n,m,f;

 30 void add(int u,int v,int flow) {

 31     e[tot] = arc(v,flow,head[u]);

 32     head[u] = tot++;

 33     e[tot] = arc(u,0,head[v]);

 34     head[v] = tot++;

 35 }

 36 void Floyd() {

 37     for(int k = 1; k <= n+n; ++k) {

 38         for(int i = 1; i <= n+n; ++i) {

 39             for(int j = 1; j <= n+n; ++j) {

 40                 if(i == j || i == k || j == k) continue;

 41                 if(!g[i][j]) g[i][j] = g[i][k]&&g[k][j];

 42             }

 43         }

 44     }

 45 }

 46 bool bfs() {

 47     queue<int>q;

 48     memset(d,-1,sizeof(d));

 49     d[S] = 1;

 50     q.push(S);

 51     while(!q.empty()) {

 52         int u = q.front();

 53         q.pop();

 54         for(int i = head[u]; ~i; i = e[i].next) {

 55             if(e[i].flow && d[e[i].to] == -1) {

 56                 d[e[i].to] = d[u] + 1;

 57                 q.push(e[i].to);

 58             }

 59         }

 60     }

 61     return d[T] > -1;

 62 }

 63 int dfs(int u,int low) {

 64     if(u == T) return low;

 65     int tmp = 0,a;

 66     for(int &i = cur[u]; ~i; i = e[i].next) {

 67         if(e[i].flow && d[e[i].to] == d[u] + 1 &&(a=dfs(e[i].to,min(e[i].flow,low)))) {

 68             e[i].flow -= a;

 69             e[i^1].flow += a;

 70             low -= a;

 71             tmp += a;

 72             if(!low) break;

 73         }

 74     }

 75     if(!tmp) d[u] = -1;

 76     return tmp;

 77 }

 78 int dinic() {

 79     int tmp = 0;

 80     while(bfs()) {

 81         memcpy(cur,head,sizeof(head));

 82         tmp += dfs(S,INF);

 83     }

 84     return  tmp;

 85 }

 86 void build(int delta) {

 87     memset(head,-1,sizeof(head));

 88     tot = 0;

 89     for(int i = 1; i <= n; ++i) {

 90         add(S,i,delta);

 91         add(i+n,T,delta);

 92     }

 93     for(int i = 1; i <= n; ++i){

 94         for(int j = n+1; j <= n+n; ++j){

 95             if(g[i][j]) add(i,j,1);

 96         }

 97     }

 98 }

 99 int main() {

100     int cs,u,v;

101     scanf("%d",&cs);

102     while(cs--) {

103         scanf("%d %d %d",&n,&m,&f);

104         S = 0;

105         T = n<<1|1;

106         memset(g,false,sizeof(g));

107         for(int i = 0; i < m; ++i) {

108             scanf("%d %d",&u,&v);

109             g[u][v+n] = true;//g[v+n][u] = true;

110             //居然是有向的。。。为什么呢？

111         }

112         for(int i = 0; i < f; ++i) {

113             scanf("%d %d",&u,&v);

114             g[u][v] = g[v][u] = true;

115         }

116         Floyd();

117         int low = 0,high = n,ans = 0;

118         while(low <= high){

119             int mid = (low + high) >>1;

120             build(mid);

121             int tmp = dinic();

122             if(tmp == n*mid){

123                 ans = mid;

124                 low = mid+1;

125             }else high = mid-1;

126         }

127         printf("%d\n",ans);

128     }

129     return 0;

130 }

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