hdu 5883 The Best Path

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

2
3 2
3
4
5
1 2
2 3
4 3
1
2
3
4
1 2
2 3
2 4





Sample Output

2

Impossible

题意：给一个图 每个点有一个值 ai 现在要找一条路径 使得 每条路径都有经过且只能经过一次，使路径上的点的 ai 异或值最大。

分析：无向欧拉图当且仅当G是连通的且没有奇度顶点。无向半欧拉图当且仅当G是连通的且恰有两个奇度顶点。

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<vector>const int maxn = 100000;const int maxm = 500000;typedef long long ll;using namespace std;int n,m,ans;int w[maxn+5],cnt[maxm+5];int main(){    int T,x,y;    scanf("%d",&T);    while(T--)    {        ans = 0;        scanf("%d %d",&n,&m);        for(int i=1; i<=n; i++) scanf("%d",&w[i]), cnt[i]=0;        for(int i=1; i<=m; i++)        {            scanf("%d %d",&x,&y);            cnt[x]++, cnt[y]++;        }        int flag = 0;        for(int i=1; i<=n; i++) if(cnt[i]%2) flag++;        if(flag>2||flag==1)        {            printf("Impossible\n");            continue;        }        for(int i=1; i<=n; i++) if((cnt[i]/2+(cnt[i]%2>0))%2) ans^=w[i];        if(flag==0)// 当奇度顶点 为0 时为欧拉图 这时要枚举 从哪个顶点为起点和终点 ，这时 这个点多经过一次。        {            int Max = 0;            for(int i=1;i<=n;i++) Max = max(Max,ans^w[i]);            ans = max(ans,Max);        }        printf("%d\n",ans);    }    return 0;}