hdu5883 The Best Path

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本文介绍了一个算法问题，目标是在特定条件下找到一条路径，使得路径上经过的每个湖泊的幸运数字通过异或操作得到的最大值。文章提供了完整的代码实现，并解释了如何使用并查集确定是否存在有效的路径，以及如何通过度数来确定可能的最大幸运数值。

The Best Path

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 798 Accepted Submission(s): 332

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

题目的意思是遍历所有边，算出每个经历过的点异或起来最大值

首先并查集算出集合数，大于1不可能；然后判欧拉回路，若度为奇的大于2,则不可能，若度为奇的等于二，则路线固定，其所经历所有点，若等于零则枚举起点

这里有个小技巧，a^a=0,0^a=a 所以异或时重复经历的点只要判经历次数是否为奇数即可

另外有个坑的是位运算优先级较低，你写

for(int i=1; i<=n; i++)
                {
                    if(ans^a[i]>mx)
                        mx=ans^a[i];
                }

就会wa，它先会判断>再异或

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define mod 10000007
#define maxn 100005

int a[maxn];
int p[maxn];
int pre[maxn];
int n,m,sz;

int fin(int x)
{
    if(x != pre[x])
        pre[x] = fin(pre[x]);
    return pre[x];
}

void unio(int x,int y)
{
    int x1=fin(x);
    int x2=fin(y);
    if(x1!=x2)
    {
        pre[x1]=x2;
        sz--;
    }
}

int main()
{
    int o,u,v;
    while(~scanf("%d",&o))
    {
        while(o--)
        {
            scanf("%d%d",&n,&m);
            for(int i=1; i<=n; i++)
            {
                scanf("%d",&a[i]);
            }
            memset(p,0,sizeof(p));
            for(int i=1; i<=n; i++)
            {
                pre[i]=i;
            }
            sz=n;

            for(int i=0; i<m; i++)
            {
                scanf("%d%d",&u,&v);
                unio(u,v);
                p[u]++;
                p[v]++;
            }

            if(sz>1)
            {
                printf("Impossible\n");
                continue;
            }

            int tot=0;
            for(int i=1; i<=n; i++)
            {
                if(p[i]%2==1)
                    tot++;
            }
            if(tot>2)
            {
                printf("Impossible\n");
            }
            else if(tot==2)
            {
                int ans=0;
                for(int i=1; i<=n; i++)
                {
                    if(((p[i]+1)/2)%2)
                        ans=ans^a[i];
                }
                printf("%d\n",ans);
            }
            else
            {
                int ans=0;
                for(int i=1; i<=n; i++)
                {
                    if(((p[i]+1)/2)%2) ans=ans^a[i];
                }
                int mx=0;
                for(int i=1; i<=n; i++)
                {
                    mx=max(mx,ans^a[i]);
                }
                printf("%d\n",mx);
            }
        }
    }
    return 0;
}