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 a_P0 XOR a_P1 XOR … XOR a_Pt. She want to make this number as large as possible. Can you help her?

参考题解
先判断图是否只有一个连通块，然后判断是否存在欧拉回路或者路径，欧拉路径每个点经过 (度数+1)/2 次，欧拉回路也一样，但是起点多走一次，每个点都可以作为起点。
欧拉通路判断条件：奇数度顶点个数最多为2个，当全部都是偶数度顶点时，图中包含欧拉回路。首先按照欧拉通路计算，计算出来的值是一定要走过的所有边构成的价值，如果存在欧拉回路，就枚举每一个顶点，因为刚刚是按照通路算的，现在遍历异或返回起点的价值。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e5+5;
int degree[maxn], value[maxn], f[maxn];

int getf(int x)
{
    return x == f[x] ? x : f[x] = getf(f[x]);
}

inline void join(int a, int b)
{
    int x = getf(a);
    int y = getf(b);
    if(x != y)  f[x] = y;
}

int main()
{
    int T, n, m;
    cin >> T;
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
        {
            degree[i] = 0;
            f[i] = i;
            scanf("%d", &value[i]);
        }
        for(int i = 1; i <= m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            degree[u]++, degree[v]++;
            join(u, v);
        }
        int sum = 0, flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(getf(i) == i)    sum++;
            if(degree[i]&1) flag++;
        }
        if(sum > 1 || flag > 2)
        {
            printf("Impossible\n");
            continue ;
        }
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            int temp = (degree[i]+1)>>1;
            while(temp--)
                ans ^= value[i];
        }
        if(flag == 0)
        {
            int temp = ans;
            for(int i = 1; i <= n; i++)
            {
                ans = max(temp^value[i], ans);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}