poj 2289 Jamie's Contact Groups【二分+最大流】【二分图多重匹配问题】

题目链接：http://poj.org/problem?id=2289

Jamie's Contact Groups

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 8473		Accepted: 2875

Description

Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend's number. As Jamie's best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend's number among the groups. Jamie takes your advice and gives you her entire contact list containing her friends' names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only one of those groups and the size of the largest group is minimized.

Input

There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend's name and the groups the friend could belong to. You can assume N is no more than 1000 and M is no more than 500. The names will contain alphabet letters only and will be no longer than 15 characters. No two friends have the same name. The group label is an integer between 0 and M - 1. After the last test case, there is a single line `0 0' that terminates the input.

Output

For each test case, output a line containing a single integer, the size of the largest contact group.

Sample Input

3 2
John 0 1
Rose 1
Mary 1
5 4
ACM 1 2 3
ICPC 0 1
Asian 0 2 3
Regional 1 2
ShangHai 0 2
0 0

Sample Output

2
2

Source

Shanghai 2004

题意：

　　N个人，M个组，给出每个人可以考虑选择的组号，要求每个人都必须选且只能选一个组，求一个划分方案使得最多人数的组的人数 是所有方案中人数最少的，结果输出这个最小值。

题解：

　　一对多的二分图多重匹配问题。网络流建图：源点到每个人连边，边权为1；每个人到可以考虑选择的组连边，边权为1；每个组到汇点连边，边权为求解的最小值。这个最小值很明显可以想到二分求解。

代码：

 #include <cstdio>
 #include <vector>
 #include <algorithm>
 #include <queue>
 #include <cstring>
 using namespace std;
 const int N = ;
 const int M = 1e6;
 const int inf = 1e9;
 int n, m, S, T;
 int dep[N], cur[N];
 int head[N];
 struct Edge{
     int v, c, nex;
     Edge(int _v=,int _c=,int _nex=):v(_v),c(_c),nex(_nex){}
 }E[M];

 int cnt;
 void add(int u, int v, int c){
     E[cnt].v = v;
     E[cnt].c = c;
     E[cnt].nex = head[u];
     head[u] = cnt++;
 }

 bool bfs() {
     queue<int> q;
     memset(dep, -, sizeof(dep));
     q.push(S); dep[S] = ;
     while(!q.empty()) {
         int u = q.front(); q.pop();
         for(int i = head[u]; ~i; i = E[i].nex) {
             int v = E[i].v;
             if(E[i].c && dep[v] == -) {
                 dep[v] = dep[u] + ;
                 q.push(v);
             }
         }
     }
     return dep[T] != -;
 }
 int dfs(int u, int flow) {
     if(u == T) return flow;
     int w, used=;
     for(int i = head[u]; ~i; i = E[i].nex) {
         int v = E[i].v;
         if(dep[v] == dep[u] + ) {
             w = flow - used;
             w = dfs(v, min(w, E[i].c));
             E[i].c -= w;  E[i^].c += w;
             if(v) cur[u] = i;
             used += w;
             if(used == flow) return flow;
         }
     }
     if(!used) dep[u] = -;
     return used;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         for(int i = ; i <= T;i++)
             cur[i] = head[i];
         ans += dfs(S, inf);
     }
     return ans;
 }
 int main() {
     char s[];
     int i, j, x, t;
     while(~scanf("%d%d", &n, &m),n+m) {
         memset(head, -, sizeof(head));
         cnt = ;
         S = n+m+; T = n+m+;
         vector<int>g[N];
         for(i = ; i <= n; ++i) g[i].clear();
         for(i = ; i < n; ++i) {
             scanf("%s", s);
             while(getchar() != '\n') {
                 scanf("%d", &x);
                 g[i].push_back(x);
             }
         }
         int l = , r = n, mid;
         while(l <= r) {
             memset(head, -, sizeof(head));
             cnt = ;
             mid = (l+r)/;
             for(i = ; i < n; ++i) {
                 for(j = ; j < g[i].size(); ++j)
                     add(i, n+g[i][j], ),add(n+g[i][j], i, );
             }
             for(i = ; i < n; ++i) add(S,i,), add(i,S,);
             for(i = n; i < n+m; ++i) add(i,T,mid),add(T,i,);
             t = dinic();
             if(t == n) r = mid - ;
             else l = mid + ;
         }
         printf("%d\n", l);
     }
 }

485ms