poj3660(Cow Contest)解题报告

Solution:

传递闭包

//if a beats b and b beats c , then a beats c

//to cow i, if all the result of content(n-1) has been known,
    //then the rank can be determined

 #include <stdio.h>
 #include <stdlib.h>
 #include <stdbool.h>
 #define maxn 1000

 //if a beats b and b beats c , then a beats c

 int main()
 {
     long n,m,i,j,k,a,b,ans,total=;
     bool vis[maxn][maxn];
     scanf("%ld%ld",&n,&m);
     for (i=;i<=n;i++)
         for (j=;j<=n;j++)
             vis[i][j]=false;
     for (i=;i<=m;i++)
     {
         scanf("%ld%ld",&a,&b);
         vis[a][b]=true;
     }
     for (i=;i<=n;i++)
         for (j=;j<=n;j++)
                 for (k=;k<=n;k++)
                     vis[j][k]=vis[j][k] || (vis[j][i] && vis[i][k]);
     //to cow i, if all the result of content(n-1) has been known,
     //then the rank can be determined
     for (i=;i<=n;i++)
     {
         ans=;
         for (j=;j<=n;j++)
             if (vis[i][j] || vis[j][i])
                 ans++;
         if (ans==n-)
             total++;
     }
     printf("%ld\n",total);
     return ;
 }

传递闭包标程+解释：

 #include <stdio.h>
 #include <stdlib.h>
 #include <stdbool.h>
 #define maxn 1000

 //n个点，点编号为1~n，m条边
 //传递闭包:判断图中任意两点是否可达
 //时间复杂度:n*n*n

 int main()
 {
     long n,m,i,j,k,a,b;
     bool vis[maxn][maxn];
     scanf("%ld%ld",&n,&m);
     for (i=;i<=n;i++)
         for (j=;j<=n;j++)
             vis[i][j]=false;
     for (i=;i<=m;i++)
     {
         scanf("%ld%ld",&a,&b);
         vis[a][b]=true;
     }
     //如果是无向图，a能到达b意味着b也能到达a，所以可以求a->b(a<b),i<j<k
     //如果没有if:操作n*n*n
     //如果有if:判断n*n+n*(n-1)*n次，操作n*(n-1)*(n-2)
     //所以直接不用if
     //从j点到k点：从j到k经过的点的编号小于等于i(i=1,2,…,n)。当然也可以从j直达到k。
     for (i=;i<=n;i++)
         for (j=;j<=n;j++)
             //if (i!=j)
                 for (k=;k<=n;k++)
                     //if (i!=j && i!=k)
                     vis[j][k]=vis[j][k] || (vis[j][i] && vis[i][k]);
                         //if (vis[j][i] && vis[i][k])
                             //vis[j][k]=true;
     for (i=;i<=n;i++)
     {
         printf("%ld : ",i);
         for (j=;j<=n;j++)
             if (vis[i][j])
                 printf("%ld ",j);
         printf("\n");
     }
     return ;
 }
 /*
 5 5
 1 2
 2 3
 1 3
 4 5
 5 3
 */

其实传递闭包跟floyd很像，原理都是一样的。