HDU 5883 欧拉路径异或值最大 水题

The Best Path

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2104    Accepted Submission(s): 841

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

解析 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

所以直接枚举取最大值就好了。

AC代码     

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int a[maxn],du[maxn],par[maxn];
int _find(int x)
{
    return x==par[x]?x:par[x]=_find(par[x]);
}
void unio(int a,int b)
{
    int ra=_find(a);
    int rb=_find(b);
    if(ra!=rb) par[rb]=ra;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            par[i]=i;du[i]=0;
            scanf("%d",&a[i]);
        }
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            unio(u,v);
            du[u]++,du[v]++;
        }
        int sum=0,flag=0;
        for(int i=1;i<=n;i++)
        {
            if(du[i]&1)sum++;
            if(par[i]==i)flag++;
        }
        if(sum>2||flag>1)
            printf("Impossible
");
        else
        {
            int ans=0;
            for(int i=1;i<=n;i++)
            {
                int temp=(du[i]+1)/2;
                while(temp)
                    ans=ans^a[i],temp--;
            }
            if(sum==0)
            {
                int temp=ans;
                for(int i=1;i<=n;i++)
                    ans=max(temp^a[i],ans);
            }
            printf("%d
",ans);
        }
    }
}