hdu 5883 The Best Path

最新推荐文章于 2020-11-25 20:40:43 发布

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The Best Path

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 801 Accepted Submission(s): 335

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

Sample Output

  
   2
Impossible

Source

2016 ACM/ICPC Asia Regional Qingdao Online

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题意：给你连通图，问存不存在一条路只经过所有的边一次，如果存在，lucky number为每经过一个点一次就异或一次的最后答案，求最大的lucky number。

思路：连通图经过所有边一次，就是欧拉路的定义。相同的数异或的话是等于0的，所以很容易可以推出每个点的贡献值为((num[i]+1)/2)%2*value[i],num为度数，value为点权值，有一种特殊情况就是形成了欧拉回路，回路的话起点的贡献值要多算一次，也就是还得异或value[i]，所以如果存在欧拉回路我们还要枚举每个点作为起点来算。下面给代码：

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
typedef long long LL;
using namespace std;
#define inf 0x3f3f3f3f
#define maxn 100005
typedef long long LL;
int value[maxn],num[maxn];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++){
            scanf("%d",&value[i]);
        }
        for(int i=0;i<m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            num[u]++;
            num[v]++;
        }
        int sum=0;
        for(int i=1;i<=n;i++){
            if(num[i]%2){
                sum++;
            }
        }
        if(sum!=0&&sum!=2){
            printf("Impossible\n");
        }
        else{
            int ans=0;
            for(int i=1;i<=n;i++){
                if(((num[i]+1)/2)%2){
                    ans^=value[i];
                }
            }
            if(!sum){
                int re=0;
                for(int i=1;i<=n;i++){
                    int now=ans^value[i];
                    re=max(now,re);
                }
                ans=re;
            }
            printf("%d\n",ans);
        }
    }
}