ZOJ 3332 Strange Country II 判断有向图联通

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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 ton. The unique way to travel in the country is taking planes.Strangely, in this strange country, for every two citiesA andB,
there is a flight from A to B or fromB 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.ThenT test cases follow. Each test case starts with a line containing an integern (0 <n
<= 100), which is the number of cities.Each of the nextn * (n - 1) / 2 lines contains 2 numbersA,
B (0 <A, B <= n, A != B), meaning that there is a flight from cityA 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

比赛的时候抓不住重点，搞不清此题的真正意图，以至于我先用mst求解，在用朱刘算法也不行。到头来，就是一道dfs深搜。汗！！！！！！

#include<stdio.h>
#include<string.h>
#define M 107
int vis[M],g[M][M],path[M];
int n,flag;
void dfs(int u,int step)
{
path[step]=u;
if(step==n){flag=1;return;}
for(int i=1;i<=n;i++)
{
if(!vis[i]&&g[u][i])
{
vis[i]=1;
dfs(i,step+1);
if(flag)return;
vis[i]=0;
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
int t;
scanf("%d",&t);
while(t--)
{
memset(g,0,sizeof(g));
memset(vis,0,sizeof(vis));
scanf("%d",&n);
int m=n*(n-1)/2,a,b;
while(m--)
{
scanf("%d%d",&a,&b);
g[a][b]=1;
}
flag=0;
for(int i=1;i<=n;i++)
{
vis[i]=1;
dfs(i,1);
if(flag)
{
for(int i=1;i<n;i++)
printf("%d ",path[i]);
printf("%d\n",path
);
break;
}
vis[i]=0;
}
if(!flag)printf("Impossible\n");
}
return 0;
}