The Best Path
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
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
【题目链接】 The Best Path
【题意】
有一个图,n个节点,m条边,每个点有一个权值,现在你需要把所有边不重复地都走一遍,求经过点的最大异或和
【思路】
能一次性走完,说明这个图有是欧拉图或半欧拉图(有欧拉回路或通路),那么根据定义可知,欧拉回路要求图中所有点的度数为偶数,欧拉通路要求图中有两个点的度数为奇数,其余均为偶数,据此可以判断出路径是否存在(注意判断图的连通性)
首先对于欧拉通路来说,起点和终点一定是奇度顶点,那么每个点经过的次数为度数除2向上取整(deg[i]+1)/2,由于异或的特殊性质,我们只要把经过次数为奇数的点的权值异或起来即可。
对于欧拉回路来说,起点会多经过一次,我们我们只要枚举起点,跟上面的结果疑惑一下,更新最大值即可
#include <cstdio>
#include <bits/stdc++.h>
#include <cmath>
#include <map>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)
typedef long long ll;
const int maxn = 100005;
const ll mod = 1e18+7;
const int INF = 1e9;
const double eps = 1e-6;
int n,m;
int pre[maxn],a[maxn];
int deg[maxn];
int find(int x)
{
int t,r=x;
while(x!=pre[x]) x=pre[x];
while(r!=x)
{
t=pre[r];
pre[r]=x;
r=t;
}
return x;
}
int main()
{
rush()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),pre[i]=i,deg[i]=0;
for(int i=0;i<m;i++)
{
int x,y;
scanf("%d%d",&x,&y);
deg[x]++,deg[y]++;
pre[find(x)]=pre[find(y)];
}
int num=0;
int odd=0;
for(int i=1;i<=n;i++)
{
if(find(i)==i) num++;
if(deg[i]&1) odd++;
}
if(num>1||!(odd==0||odd==2))
{
puts("Impossible");
continue;
}
int ans=0;
for(int i=1;i<=n;i++)
{
deg[i]=(deg[i]+1)/2;
if(deg[i]&1) ans^=a[i];
}
if(odd==0)
{
int tmp=ans;
for(int i=1;i<=n;i++) ans=max(ans,tmp^a[i]);
}
printf("%d\n",ans);
}
}