ZOJ 3332 Strange Country II

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 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

边这么多，随便跑跑就能找到路的，所以直接dfs就好了，为了输出路径所以倒着比较好。
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int maxn=1e2+10;
int T,n,x,y,vis[maxn];
vector<int> t[maxn];

bool dfs(int x,int y)
{
if (y==n){ printf("%d",x); return true;}
vis[x]=1;
for (int i=0;i<t[x].size();i++)
{
if (!vis[t[x][i]])
{
if (dfs(t[x][i],y+1))
{
printf(" %d",x); return true;
}
}
}
vis[x]=0;
return false;
}

int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
for (int i=1;i<=n;i++) t[i].clear();
for (int i=1;i+i<=n*(n-1);i++)
{
scanf("%d%d",&x,&y);
t[y].push_back(x);
}
int flag=0;
memset(vis,0,sizeof(vis));
for (int i=1;i<=n;i++)
{
if (dfs(i,1)) {flag=1; break;}
}
if (!flag) printf("Impossible");
printf("\n");
}
return 0;
}
温故知新

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
typedef long long LL;
const int low(int x) { return x&-x; }
const int mod = 1e9 + 7;
const int N = 1e2 + 10;
const int M = 1 << 14;
int T, n, m, x, y, d

, o
, v
;

bool dfs(int x, int y)
{
o[y] = x; v[x] = 1;
if (y == n) return true;
rep(i, 1, n)
{
if (v[i] || !d[x][i]) continue;
if (dfs(i, y + 1)) return true;
}
v[x] = 0;
return false;
}

int main()
{
for (inone(T); T--;)
{
inone(n); m = n*(n - 1) / 2;
ms(d, 0);
ms(v, 0);
rep(i, 1, m)
{
intwo(x, y);
d[x][y] = 1;
}
bool flag = false;
rep(i, 1, n) if (flag |= dfs(i, 1)) break;
if (flag) rep(i, 1, n) printf("%d%s", o[i], i == n ? "\n" : " ");
else puts("Impossible");
}
return 0;
}