HDU2094+拓扑排序

题意是个大坑。。。只要,保证有且只有一个人的入度是0，就yes。。

  1 #include<stdio.h>
  2 #include<string>
  3 #include<map>
  4 #include<iostream>
  5 #include<algorithm>
  6 using namespace std;
  7 const int maxn = 2005;
  8 int ind[ maxn ];
  9 struct node{
 10     int u,next;
 11 }edge[ maxn*2000 ];
 12 int cnt,head[ maxn ];
 13 map<string,int>mp;
 14 int vis[ maxn ];
 15 void init(){
 16     mp.clear();
 17     memset( vis,0,sizeof( vis ) );
 18     memset( ind,0,sizeof( ind ) );
 19     memset( head,-1,sizeof( head ) );
 20     cnt = 0;
 21 }
 22 void addedge( int a,int b ){
 23     edge[ cnt ].u = b;
 24     edge[ cnt ].next = head[ a ];
 25     head[ a ] = cnt++;
 26 }
 27 
 28 void tuopu( int n ){
 29     //printf("tuopu\n");
 30     int sum = 0;
 31     for( int i=1;i<=n;i++ ){
 32         int s = -1;
 33         for( int j=1;j<=n;j++ ){
 34             if( ind[j]==0 ){
 35                 s = j;
 36                 sum++;
 37                 ind[ j ] = -1;
 38                 break;
 39             }
 40             if( ind[ j ]<0 ){
 41                 sum = -1;
 42                 break;
 43             }
 44         }
 45         if( s==-1||sum==-1 ) break;
 46         for( int j=head[ s ];j!=-1;j=edge[ j ].next ){
 47             ind[ edge[ j ].u ]--;
 48             if( ind[ edge[ j ].u ]<0 ){
 49                 sum = -1;
 50                 break;
 51             }
 52         }
 53         if( sum==-1 ) break;
 54     }
 55     if( sum<n||sum>n ) printf("No\n");
 56     else printf("Yes\n");
 57 }
 58 
 59 int main(){
 60     int n;
 61     while( scanf("%d",&n)==1,n ){
 62         init();
 63         int cc = 1;
 64         int num1,num2;
 65         string a,b;
 66         while( n-- ){
 67             cin>>a>>b;
 68             if( mp[a]==0 ){
 69                 mp[a]=cc;
 70                 num1 = cc++;
 71             }
 72             else num1 = mp[a];
 73             if( mp[b]==0 ){
 74                 mp[b]=cc;
 75                 num2 = cc++;
 76             }
 77             else num2 = mp[b];
 78             ind[ num2 ]++;
 79             addedge( num1,num2 );
 80         }
 81         int sum = 0;
 82         //int k;
 83         for( int i=1;i<cc;i++ ){
 84             if( ind[i]==0 ){
 85                 sum++;
 86                 //k = i;
 87                 if( sum>=2 )
 88                     break;
 89             }
 90         }
 91         if( sum>=2 ){
 92             printf("No\n");
 93             continue;
 94         }
 95         if( sum==1 ){
 96             printf("Yes\n");
 97             continue;
 98         }
 99         tuopu( cc-1 );
100     }
101     return 0;
102 }
View Code

 

posted @ 2013-05-21 23:47  xxx0624  阅读(210)  评论(0)  编辑  收藏  举报