poj 2289 Jamie's Contact Groups 【二分 + 最大流】

Jamie's Contact Groups

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 6898		Accepted: 2260

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

题意：有N个人和M个小组，要求每一个人只能属于一个小组，现在已经给出每个人可以归属的小组编号（从0到M-1）。设所有小组中人数的 最多的小组所拥有的人数为IT，现在让你求最小的IT。

思路：二分枚举 + 最大流，枚举所有小组中人数最多的小组所拥有的人数，对当前枚举的mid值。建图如下：
1，超级源点到每个人引一条边，容量为1，表示一个人只能属于一个小组；
2，每个人向他所属的小组建边，容量为1；
3，所有小组向超级汇点建边，容量为mid。
最后跑一遍最大流，判断是否满流，若满流向左区间查找，反之向右区间查找。

AC代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define MAXN 1600
#define MAXM 1200000+10
#define INF 0x3f3f3f3f
using namespace std;
struct Edge
{
    int from, to, cap, flow, next;
};
Edge edge[MAXM], Redge[MAXM];
int head[MAXN], edgenum, Rhead[MAXN], Redgenum;
int dist[MAXN], cur[MAXN];
bool vis[MAXN];
int source, sink;//超级源点 超级汇点
int N, M;
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
    Edge E1 = {u, v, w, 0, head[u]};
    edge[edgenum] = E1;
    head[u] = edgenum++;
    Edge E2 = {v, u, 0, 0, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum++;
}
void getMap()
{
    source = 0, sink = N+M+1;
    char str[20];
    for(int i = 1; i <= N; i++)
    {
        addEdge(source, i, 1);//源点向每个人 建边 容量为1
        scanf("%s", str);
        int a;
        while(getchar() != '\n')
        {
            scanf("%d", &a);
            addEdge(i, a + N + 1, 1);//每个人向 所属小组建边
        }
    }
}
bool BFS(int s, int t)
{
    queue<int> Q;
    memset(dist, -1, sizeof(dist));
    memset(vis, false, sizeof(vis));
    dist[s] = 0;
    vis[s] = true;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            Edge E = edge[i];
            if(!vis[E.to] && E.cap > E.flow)
            {
                dist[E.to] = dist[u] + 1;
                vis[E.to] = true;
                if(E.to == t) return true;
                Q.push(E.to);
            }
        }
    }
    return false;
}
int DFS(int x, int a, int t)
{
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1; i = edge[i].next)
    {
        Edge &E = edge[i];
        if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap-E.flow), t)) > 0)
        {
            edge[i].flow += f;
            edge[i^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}
int Maxflow(int s, int t)
{
    int flow = 0;
    while(BFS(s, t))
    {
        memcpy(cur, head, sizeof(head));
        flow += DFS(s, INF, t);
    }
    return flow;
}
int main()
{
    while(scanf("%d%d", &N, &M), N||M)
    {
        init();
        getMap();
        memcpy(Rhead, head, sizeof(head));
        memcpy(Redge, edge, sizeof(edge));
        Redgenum = edgenum;
        int left = 0, right = N, ans = N;
        while(right >= left)
        {
            int mid = (left + right) >> 1;
            memcpy(edge, Redge, sizeof(Redge));//copy 数组
            memcpy(head, Rhead, sizeof(Rhead));
            edgenum = Redgenum;
            for(int i = 1; i <= M; i++)//枚举枚举都要建边
                addEdge(i + N, sink, mid);
            if(Maxflow(source, sink) == N)//判断是否满流
            {
                ans = min(ans, mid);
                right = mid - 1;
            }
            else
                left = mid + 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}