HDU 5883 The Best Path

The Best Path

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 852    Accepted Submission(s): 359

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 a_P0 XOR a_P1 XOR ... XOR a_Pt. 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

 

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

 

 

解析：

 

 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXN = 100000+5;
int a[MAXN];
int degree[MAXN];
int n, m;

void solve()
{
    memset(degree, 0, sizeof(degree));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    int u, v;
    while(m--){
        scanf("%d%d", &u, &v);
        ++degree[u];
        ++degree[v];
    }
    int odd_sum = 0;
    for(int i = 1; i <= n; ++i){
        if(degree[i]&1)
            ++odd_sum;
    }
    if(!(odd_sum == 0 || odd_sum == 2)){
        printf("Impossible
");
        return;
    }
    int val = 0;
    for(int i = 1; i <= n; ++i){
        if((degree[i]/2)&1)
            val ^= a[i];
    }
    if(odd_sum == 0){
        int res = 0xffffffff;
        for(int i = 1; i <= n; ++i){
            if(degree[i] != 0)
                res = max(res, val^a[i]);
        }
        printf("%d
", res);
    }
    else{
        int s = 0, e = 0;
        for(int i = 1; i <= n; ++i){
            if(degree[i]&1){
                if(s == 0){
                    s = i;
                }
                else{
                    e = i;
                    break;
                }
            }
        }
        int res = val^a[s]^a[e];
        printf("%d
", res);
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--){
        solve();
    }
    return 0;
}