2014多校8（1004）hdu4948(图论+贪心)

Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 199    Accepted Submission(s): 108
Special Judge

Problem Description

Teacher Mai has a kingdom consisting of n cities. He has planned the transportation of the kingdom. Every pair of cities has exactly a one-way road.

He wants develop this kingdom from one city to one city.

Teacher Mai now is considering developing the city w. And he hopes that for every city u he has developed, there is a one-way road from u to w, or there are two one-way roads from u to v, and from v to w, where city v has been developed before.

He gives you the map of the kingdom. Hope you can give a proper order to develop this kingdom.

 

Input

There are multiple test cases, terminated by a line "0".

For each test case, the first line contains an integer n (1<=n<=500).

The following are n lines, the i-th line contains a string consisting of n characters. If the j-th characters is 1, there is a one-way road from city i to city j.

Cities are labelled from 1.

 

Output

If there is no solution just output "-1". Otherwise output n integers representing the order to develop this kingdom.

 

Sample Input

    
    
    
    

     
     
     
     3
011
001
000
0
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     1 2 3
    
    
    
    

题意：给你一副图，没对顶点之间有且仅有一条单向的边，问你可不可以每次找出一个点，使得之前找过的顶点与它的距离不超过2，如果存在则输出点的顺序

思路：因为题目保证输入的图一定是每对顶点之间有且仅有一条单向的边，所以答案是一定存在的，做法只需要将所有顶点的入度记录下来，然后让入度从小到大的顺序输出

点的顺序即可

为什么是这样的呢，其实可以反过来想，可以倒着找一个顶点，使得当前所有的顶点到它的距离不超过2，这样等价于优先找入度最大的顶点，假设我当前找到了入度最大的顶

点u，如果存在一个离它距离大于2的顶点v，那么所有指向u的顶点以及u本身都会指向v，如果不指向v，那么一定是v指向这些顶点，那么v到u的距离就等于2了，就矛盾了，如

果都指向v，那么当前入度最大的顶点就不是u而是v了，这与u为当前最大顶点矛盾，所以这个贪心思想不会有错了，哎，这题真的不想多说什么，只是感叹贪心的题目都是无上

限的啊，想到就秒杀，想不到就= =！了