【二分图匹配】 hdu3605 Escape

Escape

http://acm.hdu.edu.cn/showproblem.php?pid=3605

Problem Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

 

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

 

Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

 

Sample Input

   
   
   
   

    
    
    
    1 1
1
1

2 2
1 0
1 0
1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES
NO
   
   
   
   

二分图的多重匹配问题，也可以用匈牙利算法解决，具体思路见代码。

#include<cstdio>
#include<cstring>
using namespace std;
#define N 100005
#define M 12
int map[N][M];//i适合在j上生存
int cap[M];//星球的容量
int num[M];//星球上人数
int link[M][N];//i星球上居住了哪些人
bool visit[M];
bool dfs(int x,int m)
{
    for(int i=0;i<m;++i)//枚举每个星球
    {
        if(!visit[i]&&map[x][i])
        {
            visit[i]=true;
            if(num[i]<cap[i])//i星球上人数小于容量
            {
                link[i][num[i]++]=x;
                return true;
            }
            for(int j=0;j<num[i];++j)
            {
                if(dfs(link[i][j],m))//给link[i][j]找新的星球
                {
                    link[i][j]=x;
                    return true;
                }
            }
        }
    }
    return false;
}
int main()
{
    int n,m;
    for(bool flag=true;~scanf("%d%d",&n,&m);flag=true)
    {
        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
                scanf("%d",&map[i][j]);
        for(int i=0;i<m;++i)
            scanf("%d",&cap[i]);
        memset(num,0,sizeof(num));
        for(int i=0;i<n;++i)//匈牙利算法
        {
            memset(visit,false,sizeof(visit));
            if(!dfs(i,m))
            {
                flag=false;
                break;
            }
        }
        if(flag)  puts("YES");
        else      puts("NO");
    }
    return 0;
}

来源： http://blog.csdn.net/ACM_Ted