hdu 4948

Kingdom

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 743    Accepted Submission(s): 347
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

析：n个城市，每一对城市之间有一条有向边连接，现在开始发展每一个城市，要求下一个要发展的城市到之前已经被发展的城市之间不超过两条边，即要么直接相连，要么中间可间隔一条边。要求输出发展顺序，不满足输出-1
每对点间都存在一条边，为竞赛图，容易想到拓扑排序，但是要逆着排序，以入度最大的为起点拓扑一下

 

#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;
const int N = 505;
int t, n, m, ans;
int in[N], res[N];
char str[N];
void solve(){
    ans = n;
    int flag = 1;
    while(flag){
        flag = 0;
        int iMax = -1;
        for(int i = 1; i <= n; i ++){
            if(iMax < in[i]){
                iMax = in[i];
                flag = 1;
            }
        }
        for(int i = 1; i <= n; i ++){
            if(iMax == in[i]){
                res[ans--] = i;
                in[i] = -2;
            }
        }
    }
}
int main(){
    while(scanf("%d", &n) && n){
        fill(in, in+n+1, 0);
        for(int i = 1; i <= n; i ++){
            scanf("%s", str+1);
            for(int j = 1; j <= n; j ++){
                t = str[j]-'0';
                if(t){
                    in[j] ++;
                }
            }
        }
        solve();
        for(int i = 1; i < n; i ++){
            printf("%d ", res[i]);
        }
        printf("%d
", res[n]);
    }
    return 0;
}