HDU1669 Jamie's Contact Groups 二分法+二分图的多重匹配

Jamie's Contact Groups
Time Limit: 15000/7000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 589    Accepted Submission(s): 211


Problem 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
2004 Asia Regional Shanghai

设limit为每组的最大人数
显然可以通过最大流/二分图多重匹配 验证能够匹配成功
二分法确定能匹配成功的最小值即可

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
#include<queue>
#include<sstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>

const int NX = 1e3+5;
const int NY = 500 + 5;

struct Edge{
    int to,next;
}edge[NX*NY];
int head[NX];

inline void addEdge(int k,int u,int v){
    edge[k].to=v;
    edge[k].next=head[u];
    head[u]=k;
}

int nx,ny;//左右集合数目
bool bmask[NY];//需找增广路时的标记数组
int vcy[NY];//vcy[i]表示右集合i顶点匹配到左集合的顶点数 i:1-ny
int cy[NY][NX];//cy[i][j]表示右集合第i个顶点 匹配的第j个顶点的标号 j从0开始
int limit;//每个右集合顶点 最多匹配的左集合顶点数

bool findPath(int u){
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].to;
        if(!bmask[v]){
            bmask[v]=1;
            if(vcy[v]<limit){
                cy[v][vcy[v]++]=u;
                return 1;
            }
            else{
                for(int j=0;j<vcy[v];++j){
                    if(findPath(cy[v][j])){
                        cy[v][j]=u;
                        return 1;
                    }
                }
            }
        }
    }
    return 0;
}

bool mulMatch(){//返回 右集合每个点最多匹配limit个点 能否完美匹配
    fill(vcy,vcy+ny+1,0);
    for(int i=1;i<=nx;++i){
        fill(bmask,bmask+ny+1,false);
        if(!findPath(i)){
            return false;
        }
    }
    return true;
}

void buildG(){
    string line,name;
    int k;//第几个分组
    int numE=0;
    fill(head,head+nx+1,-1);
    for(int i=1;i<=nx;++i){
        getline(cin,line);
        istringstream in(line);
        in>>name;
        while(in>>k){
            addEdge(numE++,i,k+1);//原题右集合从0开始 这里从1开始
        }
    }
}

int binarySearch(){
    int l=1,r=nx;
    while(l<r){
        limit=(l+r)/2;
        if(mulMatch()){
            r=limit;
        }
        else{
            l=limit+1;
        }
    }
    return r;
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    //freopen("/home/lu/Documents/w.txt","w",stdout);
    while(scanf("%d%d",&nx,&ny),nx){
        cin.ignore();
        buildG();
        printf("%d\n",binarySearch());
    }
    return 0;
}