hdu 3605 Escape (二分圖多重匹配)

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


Problem Description2012 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.
 
InputMore 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
 
OutputDetermine whether all people can live up to these stars
If you can output YES, otherwise output NO.
 
Sample Input

1 1112 21 01 01 1

 
Sample Output

YESNO

 
Source2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

多重匹配即 X集合上的點對應 Y集合上多個點而 Y集合上的點對應 X中的一個點.

可以用一個二維數組記錄匹配的對象link[M][N]，再用一個數組cnt[M]記錄該點已經匹配幾個點了，是否超出最大匹配，若超出最大匹配，則在該點已經匹配的點中尋找增廣路徑。

若某個點不能匹配成功，則尋找失敗。。

#include"stdio.h"#include"string.h"#define N 100005#define M 15bool g[N][M],vis[M];          //存邊的權值0、1，記錄是否訪問過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;}