ZOJ 3332 Strange Country II

Strange Country II


Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge


You want to visit a strange country. There are n cities in the country. Cities are numbered from 1 to n. The unique way to travel in the country is taking planes. Strangely, in this strange country, for every two cities A and B, there is a flight from A to B or from B to A, but not both. You can start at any city and you can finish your visit in any city you want. You want to visit each city exactly once. Is it possible?

Input

There are multiple test cases. The first line of input is an integer T (0 < T <= 100) indicating the number of test cases. Then T test cases follow. Each test case starts with a line containing an integer n (0 < n<= 100), which is the number of cities. Each of the next n * (n - 1) / 2 lines contains 2 numbers AB (0 < AB <= nA != B), meaning that there is a flight from city A to city B.

Output

For each test case:

  • If you can visit each city exactly once, output the possible visiting order in a single line please. Separate the city numbers by spaces. If there are more than one orders, you can output any one.
  • Otherwise, output "Impossible" (without quotes) in a single line.

Sample Input

3
1
2
1 2
3
1 2
1 3
2 3

Sample Output

1
1 2
1 2 3
dfs#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int a[105][105];
int ans[105];
int vis[105];
int n;
bool res;
void dfs(int x,int cnt)
{
	if(cnt>n)
		return;
	if(res)
	{
		ans[cnt]=x;
		return;
	}
    if(cnt==n)
	{
		res=true;
		ans[cnt]=x;
		return;

	}
    for(int i=1;i<=n;i++)
	{
		if(a[x][i]&&!vis[i])
		{
			vis[i]=1;
			dfs(i,cnt+1);
			vis[i]=0;
			if(res)
			{
				ans[cnt]=x;
				return;
			}

		}
	}
}
int main()
{
	int t;
	int x,y;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(a,0,sizeof(a));
		for(int i=1;i<=n*(n-1)/2;i++)
		{
			scanf("%d%d",&x,&y);
			a[x][y]=1;
		}
		memset(vis,0,sizeof(vis));
		res=false;
		for(int i=1;i<=n;i++)
		{
			vis[i]=1;
			dfs(i,1);
			vis[i]=0;
			if(res)
				break;
		}
		if(!res)
		{
           printf("Impossible\n");
		   continue;
		}
		for(int i=1;i<=n;i++)
		{
			if(i!=n)
			printf("%d ",ans[i]);
			else
				printf("%d",ans[i]);
		}
		printf("\n");
	}
	return 0;
}

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

扫码关注云+社区

领取腾讯云代金券