hdu 3605 二分图多重匹配

和上一题大同小异，不过数据稍微强点。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 using namespace std;
  5 
  6 const int X = 100000;
  7 const int Y = 10;
  8 int head[X];
  9 int sz[Y];
 10 int cnt[Y];
 11 int mark[Y][X];
 12 bool visit[Y];
 13 int e;
 14 
 15 void init()
 16 {
 17     e = 0;
 18     memset( head, -1, sizeof(head) );
 19 }
 20 
 21 struct Edge 
 22 {
 23     int v, next;
 24 } edge[X * Y];
 25 
 26 void addEdge( int u, int v )
 27 {
 28     edge[e].v = v;
 29     edge[e].next = head[u];
 30     head[u] = e++;
 31 }
 32 
 33 int dfs( int u )
 34 {
 35     for ( int i = head[u]; i != -1; i = edge[i].next )
 36     {
 37         int v = edge[i].v;
 38         if ( !visit[v] )
 39         {
 40             visit[v] = 1;
 41             if ( cnt[v] < sz[v] )
 42             {
 43                 mark[v][cnt[v]++] = u;
 44                 return 1;
 45             }
 46             else
 47             {
 48                 for ( int j = 0; j < cnt[v]; j++ )
 49                 {
 50                     if ( dfs( mark[v][j] ) )
 51                     {
 52                         mark[v][j] = u;
 53                         return 1;
 54                     }
 55                 }
 56             }
 57         }
 58     }
 59     return 0;
 60 }
 61 
 62 int main ()
 63 {
 64     int n, m;
 65     while ( scanf("%d%d", &n, &m) != EOF )
 66     {
 67         init();
 68         for ( int i = 0; i < n; i++ )
 69         {
 70             for ( int j = 0; j < m; j++ )
 71             {
 72                 int sn;
 73                 scanf("%d", &sn);
 74                 if ( sn )
 75                 {
 76                     addEdge( i, j );
 77                 }
 78             }
 79         }
 80         for ( int i = 0; i < m; i++ )
 81         {
 82             scanf("%d", sz + i);
 83         }
 84         memset( cnt, 0, sizeof(cnt) );
 85         bool flag = true;
 86         for ( int i = 0; i < n; i++ )
 87         {
 88             memset( visit, 0, sizeof(visit) );
 89             if ( !dfs(i) )
 90             {
 91                 flag = false;
 92                 break;
 93             }
 94         }
 95         if ( flag )
 96         {
 97             puts("YES");
 98         }
 99         else
100         {
101             puts("NO");
102         }
103     }
104     return 0;
105 }

 

posted @ 2015-08-22 23:21  hxy_has_been_used  阅读(132)  评论(0)  编辑  收藏  举报