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 }

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