ZOJ 3332 Strange Country II



K - Strange Country II

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3332

Description

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 Bto 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

3121 231 21 32 3

Sample Output

11 21 2 3

题目大意：给你n*（n-1）条路线，代表可以从a走到b。

问你遍历所有点的路径输出。（可以输出任意解，每个城市只能走一次）

因为可以输出任意解，直接暴力搜索就好了。

#include <cstdio>#include <algorithm>#include <map>#include <cstring>#include <cmath>#include <iostream>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1#define LL __int64typedef long long ll;#define PI 3.1415926int mat[150][150];int vis[150];int b[150];int n,l,flag;void dfs(int x){    int i;    if(l==n)    {        flag=1;        return;    }    for(i=1; i<=n; i++)    {        if(!vis[i]&&mat[x][i])        {             b[l++]=i;            vis[i]=1;            dfs(i);            if(flag==1)                return;            vis[i]=0;            l--;        }    }    return;}int main(){    int t,i;    cin>>t;    while(t--)    {        int a,c;        cin>>n;        if(n==1)        {            cout<<1<<endl;            continue;        }        memset(b,0,sizeof(b));        memset(mat,0,sizeof(mat));        memset(vis,0,sizeof(vis));        for(i=1; i<=n*(n-1)/2; i++)        {            cin>>a>>c;            mat[a][c]=1;        }        flag=0;        for(i=1; i<=n; i++)        {            l=0;            b[l++]=i;            vis[i]=1;            dfs(i);            l--;            if(flag)                break;                vis[i]=0;        }        if(flag)        {            for(i=0; i<n; i++)            {                if(i!=n-1)                    cout<<b[i]<<" ";                else                    cout<<b[i];            }            cout<<endl;        }        else            cout<<"Impossible"<<endl;    }    return 0;}