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
題目鏈接：HDU⑸883
題目大意： n 個點 m 條無向邊的圖，找1個歐拉通路/回路使得這個路徑所有結(jié)點的異或值最大
題目思路：根據(jù)歐拉路的性質(zhì)，
在無向圖中，
歐拉通路：兩個點度數(shù)為奇數(shù)，其余點度數(shù)為偶數(shù)
歐拉回路：所有點度數(shù)為偶數(shù)
所以分為兩種情況討論：
1.歐拉回路
出發(fā)點也是終點，所以要遍歷哪一個點為出發(fā)點（由于出發(fā)點多亦或1次，出發(fā)點不同結(jié)果不同）
2.歐拉通路
兩個奇數(shù)的點已知，肯定為出發(fā)點和終點。
注意：
1. 由于要經(jīng)過的是每條河而不是每一個湖，所以，只需要判斷有河經(jīng)過的湖是不是聯(lián)通。
2. 存在河u - > v（u == v） 這些點也要處理，由于可能存在這些點是孤立的，但是也要經(jīng)過。這里wa了好久
以下是代碼：
#include <iostream>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <functional>
#include <numeric>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <deque>
#include <list>
using namespace std;
#define LL long long
int a[100005];
int d[100005];
int cc[100005];
#define MAXN 100005
int fa[MAXN] = {0};
int ranks[MAXN] = {0};
void initialise(int n) //初始化
{
for (int i = 1; i <= n; i++)
fa[i] = i,ranks[i] = 1;
}
int getfather(int v) //父節(jié)點
{
return (fa[v] == v) ? v : fa[v] = getfather(fa[v]);
}
void merge(int x,int y) //合并
{
x = getfather(x);
y = getfather(y);
if (x != y)
fa[x] = y,ranks[y] += ranks[x];
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
memset(d,0,sizeof(d));
memset(cc,0,sizeof(cc));
for (int i = 1; i <= n; i++)
{
scanf("%d",&a[i]);
}
initialise(n);
for (int i = 0; i < m; i++)
{
int u,v;
scanf("%d%d",&u,&v);
cc[u] = 1;
cc[v] = 1;
d[u]++;
d[v]++;
merge(u,v);
}
int cnt = 0;
for (int i = 1; i <= n; i++)
{
if (fa[i] == i && cc[i])
{
cnt++;
}
}
if (cnt != 1)
{
printf("Impossible\n");
continue;
}
cnt = 0;
long long ans = 0;
for (int i = 1; i <= n; i++)
{
if (d[i] % 2 == 0)
{
int kk = d[i] / 2;
if(kk % 2)
{
ans ^= a[i];
}
}
else
{
if (((d[i] + 1) / 2) % 2) ans ^= a[i];
cnt++;
}
}
if(cnt == 0)
{
ans = ans ^ a[1];
for (int i = 2; i <= n; i++)
{
ans = max(ans ^ a[i],ans);
}
}
if (cnt == 0 || cnt == 2)printf("%lld\n",ans);
else printf("Impossible\n");
}
return 0;
}
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