xfstart07
Get busy living or get busy dying

#include < iostream >
using   namespace  std;

int  n;
bool  a[ 1010 ][ 1010 ];
bool  f[ 1010 ][ 1010 ];
int  Win[ 501 ];
int  main()
{
    scanf(
" %d " , & n);
    
for ( int  i = 1 ;i <= n; ++ i)
        
for ( int  j = 1 ;j <= n; ++ j){
            
int  c;
            scanf(
" %d " , & c);
            a[i][j]
= a[i + n][j] = a[i][j + n] = a[i + n][j + n] =  c == 1 ;
        }
    
for ( int  i = 1 ;i <= 2 * n; ++ i)
        f[i][i
+ 1 ] = 1 ;
    
for ( int  k = 3 ;k <= n + 1 ; ++ k)
        
for ( int  i = 1 ;i <= 2 * n - k + 1 ; ++ i){
            
int  j = i + k - 1 ;
            
for ( int  t = i + 1 ;t < j; ++ t)
                
if (f[i][t] && f[t][j] && (a[i][t] || a[j][t])){
                    f[i][j]
= 1 ;
                    
break ;
                }
        }
    
int  l = 0 ;
    
for ( int  i = 1 ;i <= n; ++ i)
        
if (f[i][i + n])
            Win[
++ l] = i;
    printf(
" %d\n " ,l);
    
for ( int  i = 1 ;i <= l; ++ i)
        printf(
" %d\n " ,Win[i]);
    
return   0 ;
}




posted on 2009-05-03 10:09 xfstart07 阅读(121) 评论(0)  编辑 收藏 引用 所属分类: 代码库

只有注册用户登录后才能发表评论。
网站导航: 博客园   IT新闻   BlogJava   知识库   博问   管理