Zoj3332-Strange Country II(有向竞赛图)

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 A, B (0 < A, B <= n, A != 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

解析: 裸的有向竞赛图，有向竞赛图要满足的条件是有N*(N-1)/2条边，也就是每个点与其他点都有边相连，求哈密顿通路。

代码

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
const int maxn=;
int N,ans[maxn];
bool link[maxn][maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        memset(link,false,sizeof(link));
        int tot=N*(N-)/;
        int u,v;
        for(int i=;i<=tot;i++)
        {
            scanf("%d%d",&u,&v);
            link[u][v]=true;
        }
        ans[]=;
        int i,j,k;
        for(k=;k<=N;k++)
        {
            for(j=;j<k;j++) if(link[k][ans[j]]) break;
            for(i=k;i>j;i--) ans[i]=ans[i-];
            ans[j]=k;
        }
        for(int i=;i<=N;i++) printf("%d%c",ans[i],i==N?'\n':' ');
    }
    return ;
}