ZOJ 3332 Strange Country II

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3757

题面：  

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 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

解题：

    数据量比较小，直接dfs就可以了，加一个flag标记及时返回。因为不小心忽略了每一个只访问一次这个条件，wa了一次。

代码：

#include <iostream>
#include <cstdio>
#include <cstring> 
using namespace std;
bool map[105][105];
bool vis[105],flag;
int n,t,x,a,b;
int road[105],cnt=0;
void dfs(int x)
{
	//及时返回 
	if(flag)return;
	road[cnt++]=x;
	vis[x]=1;
	bool sign=true;
	//检验是否找到 
	for(int i=1;i<=n;i++)
	{
		if(vis[i]==0)
		{
			sign=false;
			break;
		}
	}
	//已经找到 
	if(sign)
	{
		flag=true;
		return;
	}
	else
	{
		for(int i=1;i<=n;i++)
		{
			//注意该点要未被访问过 
			if(map[x][i]&&!vis[i])
			{
			  dfs(i);
			  //注意还原 
			  vis[i]=0;
			  cnt--;
			}
		}
	}
}
int main()
{
	
	scanf("%d",&t);
	while(t--)
	{
		flag=false;
		memset(map,0,sizeof(map));
		scanf("%d",&n);
		x=n*(n-1)/2;
		//读入操作 
		for(int i=0;i<x;i++)
		{
		  scanf("%d%d",&a,&b);
		  map[a][b]=1;	
		}
		for(int i=1;i<=n&&!flag;i++)
		{
			cnt=0;
			memset(vis,0,sizeof(vis));
			dfs(i);
		}
		if(flag)
		{
			printf("%d",road[0]);
			for(int i=1;i<n;i++)
			printf(" %d",road[i]);
			printf("\n");
		}
		else printf("Impossible\n");
	}
	return 0;
}