1007 The Best Path（无向图的欧拉路）

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分类专栏： acm 2016 青岛网络赛

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探讨了如何寻找一条经过所有边恰好一次的欧拉路径，并在此基础上计算最大点权亦或值。介绍了解决方案的算法思路及实现细节。

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The Best Path

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 541 Accepted Submission(s): 220

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

这题让我对欧拉路有了深一点的认识。

欧拉路：分有向图的欧拉路和无向图的欧拉路。前几天做过有向图的欧拉路，这题就是无向图的欧拉路（后面也是针对无向图来讲）。

不管是无向图也好，有向图也罢，欧拉路分为点欧拉和边欧拉。点欧拉就是每个点只能走一次，边欧拉就是每条边只能走一次。

这题是边欧拉，同时，每个点都要走一次，要你输出走这条欧拉路的最大点权的亦或值

边欧拉，两种情况：1.每个点的度数都是偶数，这时，不论从哪个点出发，都能走出欧拉路，且最后一定会回到出发点，也就是说起点会经历奇数次（（度+1）/2）通过，其它点会经历偶数次（度/2）通过（所以这题如果是这种情况，就只要考虑起点是哪一个就ok，因为同一个点偶数次亦或就是1） 2.有两个点的度是奇数，那么起点必定在这两个点中，无论哪个点都可以作为起点，他们会经历（（度+1）/2）次（刚开始在那个点也算经历一次）。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;
int T, val[100005], du[100005], pre[100005];
int find(int x){return pre[x] == x ? x : pre[x] = find(pre[x]);}
int main()
{
    cin >> T;
    while(T--){
        memset(du, 0, sizeof(du));
        int n, m, flag = 0, odd = 0;
        cin >> n >> m;
        for(int i = 1; i <= n; i++) pre[i] = i;
        for(int i = 1; i <= n; i++) scanf("%d", &val[i]);
        for(int i = 1; i <= m; i++){
            int u, v;
            scanf("%d %d", &u, &v);
            du[u]++; du[v]++;
            int x = find(u), y = find(v);
            if(x != y) pre[x] = y;
        }
        for(int i = 1; i <= n; i++){
            if(pre[i] == i) flag++;
            if(du[i] % 2 != 0) odd++;
        }
        if(flag > 1) { printf("Impossible\n"); continue;}
        if(odd == 2){
            int ans = 0;
            for(int i = 1; i <= n; i++){
                du[i] = (du[i] + 1) / 2;
                if(du[i] & 1)
                    ans ^= val[i];
            }
            printf("%d\n", ans);
        }
        else if(odd == 0){
            int ans = 0;
            for(int i = 1; i <= n; i++)
                ans = max(ans, ans ^ val[i]);
            printf("%d\n", ans);
        }
        else printf("Impossible\n");
    }
    return 0;
}