poj 3894 System Engineer (二分圖最大匹配--匈牙利算法)

System Engineer
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 507   Accepted: 217

Description

Bob is a skilled system engineer. He is always facing challenging problems, and now he must solve a new one. He has to handle a set of servers with differing capabilities available to process job requests from persistent sources - jobs that need to be processed over a long or indefinite period of time. A sequence of persistent job requests arrives revealing a subset of servers capable of servicing their request. A job is processed on a single server and a server processes only one job. Bob has to schedule the maximum number of jobs on the servers. For example, if there are 2 jobs j1, j2 and 2 servers s1, s2, job j1 requiring the server s1, and job j2 requiring also the server s1. In this case Bob can schedule only one job. Can you help him?
In the general case there are n jobs numbered from 0 to n-1, n servers numbered from n to 2*n-1, and a sequence of job requests. The problem asks to find the maximum number of jobs that can be processed.
Input

The program input is at most 1 MB. Each data set in the file stands for a particular set of jobs. A data set starts with the number n (n <= 10000) of jobs, followed by the list of required servers for each job, in the format:jobnumber: (nr_servers) s1 ... snr_servers The program prints the maximum number of jobs that can be processed.
White spaces can occur freely in the input. The input data are correct and terminate with an end of file.
Output

For each set of data the program prints the result to the standard output from the beginning of a line.
Sample Input

2
0: (1) 2
1: (1) 2
1
0: (1) 1Sample Output

1
1Hint

There are two data sets. In the first case, the number of jobs n is 2, numbered 0 and 1. The sequence of requests for job 0 is: 0: (1) 2, meaning that job 0 requires 1 sever, the server numbered 2. The sequence of requests for job 1 is: 1: (1) 2, meaning that job 1 requires 1 sever, the server numbered 2. The result for the data set is the length of the maximum number of scheduled jobs, 1.
Source

Southeastern European Regional Programming Contest 2009

題意：分配工作。一個工作和一個人一一對應。

代碼：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

struct edge
{
    int to,next;
}e[100000];
int index;

bool visit[10010];   //記錄v2中的某個點是否被搜索過
int match[10010];    //記錄與V2中的點匹配的點的編號
int head[10010];
int m,cnt;   

//向圖中加邊，注意加入的是有向邊
//u為v的後繼節點即v------>u
void addedge(int u,int v)
{
    e[index].to=v;
    e[index].next=head[u];
    head[u]=index;
    index++;
}

//匈牙利算法(鄰接表存圖)
bool dfs(int u)
{
    int i,v;
    for(i=head[u];i;i=e[i].next)
    {
        v=e[i].to;
        if(!visit[v])    //如果節點u與v相鄰且未被查找過
        {
            visit[v]=true;    //標記v已查找過
            if(match[v]==-1||dfs(match[v]))    //如果v未在前一個匹配M中,或者v在匹配M中，但是
            {                                  //從與v相鄰的節點出發可以有增廣路徑
                match[v]=u;             //記錄查找成功，更新匹配即"取反"
                return true;            //返回查找成功。
            }
        }
    }
    return false;
}

void init()
{
	int aa,k,bb,i;
	memset(head,0,sizeof(head));      //切記要初始化
    memset(e,0,sizeof(e));
    index=1;
	for(i=1;i<=m;i++)
    {
        scanf("%d: (%d)",&aa,&k);
        //printf("ok: %d %d\n",aa,k);
        while(k--)
        {
            scanf("%d",&bb);
            bb=bb-cnt+1;
            addedge(bb,aa+1);
        }
    }
}

int main()
{
    int i;
    while(scanf("%d",&cnt)!=EOF)
    {
        m=cnt;
        init();
        int ans=0;
        memset(match,-1,sizeof(match));
        for(i=1;i<=cnt;i++)
        {
            memset(visit,0,sizeof(visit));    //清空上次搜索時的標記
            if(dfs(i))                 //從i節點嘗試擴展
               ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}