2181 哈密顿绕行世界问题
好吧，没什么难度的，只是要注意那个 ‘0’ 的判断我用了控制台函数……这个不会的可以去查资料，百度之，谷歌之……
/* THE PROGRAM IS MADE BY PYY *//*----------------------------------------//Copyright (c) 2011 panyanyany All rights reserved.URL : http://acm.hdu.edu.cn/showproblem.php?pid=2181Name : 2181 哈密顿绕行世界问题Date :Friday，August 12，2011Time Stage : 1 hours aroundResult:44004102011-08-12 21:04:56 Accepted 2181 0MS 192K 1475 B C++ pyyTest Data:Review://----------------------------------------*/#include <stdio.h>#include <string.h>#define max(a, b) (((a) > (b)) ? (a) : (b))#define min(a, b) (((a) < (b)) ? (a) : (b))#define infinity0x7f7f7f7f#define minus_inf0x80808080#define MAXSIZE 21int way, m ;int city[MAXSIZE][3], route[MAXSIZE], used[MAXSIZE] ;void dfs (int c, int cnt){int i ;route[cnt] = c ;if (c == m && cnt == 20){printf ("%d: %d ", way++, m) ;for (i = 1 ; i < 20 ; ++i)printf ("%d ", route[i]) ;printf ("%d\n", route[i]) ;}for (i = 0 ; i < 3 ; ++i){if (!used[city[c][i]]){used[city[c][i]] = 1 ;dfs (city[c][i], cnt + 1) ;used[city[c][i]] = 0 ;}}}int main (){char c ;int i ;while (1){c = getc (stdin) ;// 从输入流中提取一个字符if (c == '0')break ;ungetc (c, stdin) ; // 把这个字符还回去for (i = 1 ; i <= 20 ; ++i)scanf ("%d%d%d", &city[i][0], &city[i][1], &city[i][2]) ;scanf ("%d", &m) ;memset (used, 0, sizeof (used)) ;way = 1 ;for (i = 0 ; i < 3 ; ++i){used[city[m][i]] = 1 ;dfs (city[m][i], 1) ;used[city[m][i]] = 0 ;}getchar () ; // 别忘了吃掉一个多余的回车键}return 0 ;}