Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 500    Accepted Submission(s): 231
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

Author

xudyh

Source

2014 Multi-University Training Contest 8

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<map>
using namespace std;
int n,mp[505][505];
struct node
{
    int x,cnt;
}e[505];
bool cmp(node a,node b)
{
    return a.cnt<b.cnt;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%1d",&mp[i][j]);
            }
            e[i].x=i;
            e[i].cnt=0;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(mp[i][j])
                {
                    e[j].cnt++;
                }
            }
        }
        sort(e,e+n,cmp);
        printf("%d",e[0].x+1);
        for(int i=1;i<n;i++)
        {
            printf(" %d",e[i].x+1);
        }
        printf("\n");
    }
    return 0;
}