Codeforces Round #738 (Div. 2) C. Mocha and Hiking

传送门

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><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><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;
		}
	}
}

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Origin blog.csdn.net/p15008340649/article/details/119726875