题意:
给你一个有向图,问你是否存在经过所有点一次且仅一次的路径。
思路:
对于一般情况,我们只需要考虑是否存在 < i , n + 1 > 和 < n + 1 , i + 1 > <i,n+1>和<n+1,i+1> <i,n+1>和<n+1,i+1>这样的路径,存在,则有解。除此之外,还有两种特殊情况,即当 < n , n + 1 > 或 < n + 1 , 1 > <n,n+1>或<n+1,1> <n,n+1>或<n+1,1>存在时,也是有解的。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<math.h>
using namespace std;
#define ll long long
#define mod 998244353ll
int a[10010];
int in[10010];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i = 1; i <= n; i++)
{
cin>>a[i];
}
int minn = 0, flag = 0;
for(int i = 1; i <= n; i++)
{
if(i == n && !a[i])
{
flag = 1;
}
if(i == 1 && a[i])
{
flag = 2;
break;
}
if(!a[i])
{
if(a[i+1])
{
minn = i;
break;
}
}
}
if(minn)
{
for(int i = 1; i <= minn; i++)cout<<i<<" ";
cout<<n+1<<" ";
for(int i = minn+1; i <= n; i++)cout<<i<<" ";
cout<<endl;
}
else if(flag == 1)
{
for(int i = 1; i <= n+1; i++)cout<<i<<" ";
cout<<endl;
}
else if(flag == 2)
{
cout<<n+1<<" ";
for(int i = 1; i <= n; i++)
cout<<i<<" ";
cout<<endl;
}
else
{
cout<<-1<<endl;
}
}
}