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




/*欧拉路的判断，
寻找最多的幸运数;
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int big = 1e5 + 7;
int a[big];//每个节点的度数;
int  b[big];//湖的编号;
int z[big], e, f, flag;
bool judge(int n)//判断是否为欧啦路;
{
    int sum = 0, h;
    for(int i = 1; i <= n; i ++)
        if(z[i] == i) sum ++;
    if(sum != 1) return false;
    sum = flag = 0;
    for(int i = 1; i <= n; i ++)
    {
        if(a[i]&1)
        {
            sum ++;
            a[i] ++;
        }
        h = a[i]/2;
        if(!(h&1)) b[i] = 0;
        }
    if(sum != 0 && sum != 2) return false;
    if(sum == 2) flag = 1;
    return true;
}
void G(int n)//湖的标号;
{
    for(int i = 1; i <= n; i ++)
        z[i] = i;
}
int Find(int key)
{
    int s = key;
    while(s != z[s])
        s = z[s];
    int j = key, e;
    while(j != s)
    {
        e = z[j];
        z[j] = s;
        j = e;
    }
    return s;
}
int main()
{
    int t, n, m, u, v, p, q;
    scanf("%d", &t);
    while(t --)
    {
        scanf("%d %d", &n, &m);
        memset( a, 0, sizeof( a));
        for(int i = 1; i <= n; i ++)
            scanf("%d", &b[i]);
        G(n);//湖的编号;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d %d", &u, &v);
            a[u] ++;
            a[v] ++;
            p = Find(u);
            q = Find(v);
            if(p == q) continue;
            if(p > q) swap( p, q);
            z[q] = p;
        }
        if(judge(n))
        {
            int temp = b[1], ans = 0;
            for(int i = 2; i <= n; i ++)
                temp = temp xor b[i];
            if(flag) ans = temp;
            else
            {
                for(int i = 1; i <= n; i ++)
                    ans = max( ans, temp xor b[i]);
            }
            printf("%d\n", ans);
        }
        else printf("Impossible\n");
    }
    return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int maxn = 112345;
int arr[maxn];
int fnd(int x){
    return x == arr[x] ? x : arr[x] = fnd(arr[x]);
}
void join(int x,int y){
    x = fnd(x),y = fnd(y);
    if(x != y)
        arr[x] = y;
}
int val[maxn],ind[maxn];
int cal(int n){
    int bio = 0,odd = 0;
    for(int i=1;i<=n;i++){
        if(arr[i] == i && ind[i] != 0)
            bio++;
        odd += ind[i] % 2;
    }
    if(bio >= 2)
        return -1;
    if(bio == 0){
        return *max_element(val+1,val+1+n);
    }
    if(odd == 0){
        int all = 0,ret = 0;
        for(int i=1;i<=n;i++){
            if((ind[i]/2) & 1) all ^= val[i];
        }
        for(int i=1;i<=n;i++){
            if(ind[i])
                ret = max(ret,all ^ val[i]);
        }
        return ret;
    }
    if(odd == 2){
        int ret = 0;
        for(int i=1;i<=n;i++){
            if(ind[i]){
                if(ind[i] % 2){
                    if(((ind[i]+1)/2)&1) ret ^= val[i];
                }
                else{
                    if((ind[i]/2)&1) ret ^= val[i];
                }
            }
        }
        return ret;
    }
    return -1;
}


int main(){
    int T;
    scanf("%d",&T);
    int n,m;
    while(T-- && ~scanf("%d %d",&n,&m)){
        memset(ind,0,sizeof(ind));
        for(int i=1;i<=n;i++){
            scanf("%d",&val[i]);
            arr[i] = i;
        }
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d %d",&x,&y);
            join(x,y);
            ind[x]++,ind[y]++;
        }
        int cnt = cal(n);
        if(cnt == -1){
            puts("Impossible");
        }
        else{
            printf("%d\n",cnt);
        }
    }
    return 0;
}