hdu 5883 The Best Path 2016ACM/ICPC青岛赛区网络赛1006

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本文探讨了一种特殊的旅行路径问题，在此问题中，旅行者需要通过每条河流恰好一次来寻找一条路径，并且该路径的幸运数尽可能大。文章提供了具体的输入输出示例及算法实现。

Problem Description

Alice is planning her travel route in a beautiful valley. In this valley, there are

N lakes, and

M rivers linking these lakes. Alice wants to start her trip from one lake, and enjoys the landscape by boat. That means she need to set up a path which go through every river exactly once. In addition, Alice has a specific number (

a1,a2,...,an) for each lake. If the path she finds is

P0→P1→...→Pt, the lucky number of this trip would be

aP0XORaP1XOR...XORaPt. She want to make this number as large as possible. Can you help her?

Input

The first line of input contains an integer

t, the number of test cases.

t test cases follow.

For each test case, in the first line there are two positive integers

N (N≤100000) and

M (M≤500000), as described above. The

i-th line of the next

N lines contains an integer

ai(∀i,0≤ai≤10000) representing the number of the

i-th lake.

The

i-th line of the next

M lines contains two integers

ui and

vi representing the

i-th river between the

ui-th lake and

vi-th lake. It is possible that

ui=vi.

Output

For each test cases, output the largest lucky number. If it dose not have any path, output "Impossible".

Sample Input

Sample Output

2
Impossible

因为要求每条边都只走一遍。所以这是个一笔画问题

度为奇数的点>2则不可行

若度为奇数的点=2

那么一定是一个奇点入一个奇点出

首位次数为(deg+1)/2,中间点次数为deg/2

若度为奇数的点=0

那么一定是同一个点作为起点和终点

其他点次数为deg/2

然后枚举起点，起点计算次数多一次，最后取最大值即可

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int indeg[100001];
int a[100001],s[1000001],t[1000001];
int main()
{
	int T;
	scanf("%d",&T);
	while(T>0)
	{
		T--;
		int n,m;
		scanf("%d%d",&n,&m);
		int i,j;
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		memset(indeg,0,sizeof(indeg));
		for(i=1;i<=m;i++)
		{
			scanf("%d%d",&s[i],&t[i]);
			indeg[s[i]]++;
			indeg[t[i]]++;
		}
		int sum=0;
		for(i=1;i<=n;i++)
			sum+=indeg[i]%2;
		if(sum>2)
			printf("Impossible\n");
		else
		{
			if(sum==0)
			{
				int ans=0;
				for(i=1;i<=n;i++)
				{
					for(j=1;j<=indeg[i]/2;j++)
						ans^=a[i];
				}
				int maxx=0;
				for(i=1;i<=n;i++)
					maxx=max(maxx,(ans^a[i]));
				printf("%d\n",maxx);
			}
			else
			{
				int ans=0;
				for(i=1;i<=n;i++)
				{
					if(indeg[i]%2==0)
					{
						for(j=1;j<=indeg[i]/2;j++)
							ans^=a[i];
					}
					else
					{
						for(j=1;j<=indeg[i]/2+1;j++)
							ans^=a[i];
					}
				}
				printf("%d\n",ans);
			}
		}
	}
	return 0;
}