首页 > ACM题库 > HDU-杭电 > hdu 3605-escape-图-[解题报告]hoj
2014
11-27

hdu 3605-escape-图-[解题报告]hoj

Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4298    Accepted Submission(s): 1129



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
 


Source

多重匹配即 X集合上的点对应 Y集合上多个点而 Y集合上的点对应
X
中的一个点.

可以用一个二维数组记录匹配的对象link[M][N],再用一个数组cnt[M]记录该点已经匹配几个点了,是否超出最大匹配,若超出最大匹配,则在该点已经匹配的点中寻找增广路径。

若某个点不能匹配成功,则查找失败。。

#include"stdio.h"
#include"string.h"
#define N 100005
#define M 15
bool g[N][M],vis[M];          //存边的权值0、1,记录是否访问过
int lim[M],cnt[M],link[M][N];  
int n,m;
int find(int k)
{
    int i,j;
    for(i=0;i<m;i++)
    {
        if(!vis[i]&&g[k][i])
        {
            vis[i]=1;
            if(cnt[i]<lim[i])
            {
                link[i][cnt[i]++]=k;
                return 1;
            }
            for(j=0;j<cnt[i];j++)
            {
                if(find(link[i][j]))
                {
                    link[i][j]=k;
                    return 1;
                }
            }
        }
    }
    return 0;
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        memset(g,0,sizeof(g));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%d",&g[i][j]);
            }
        }
        for(i=0;i<m;i++)
        {
            scanf("%d",&lim[i]);
        }
        memset(link,0,sizeof(link));
        memset(cnt,0,sizeof(cnt));
        for(i=0;i<n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(!find(i))
                break;
        }
        if(i==n)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

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  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。