hdu5883 The Best Path 2016-09-21 21:31 92人阅读 评论(0) 收藏

The Best Path

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 798    Accepted Submission(s): 332

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

 

题目的意思是遍历所有边，算出每个经历过的点异或起来最大值

首先并查集算出集合数，大于1不可能；然后判欧拉回路，若度为奇的大于2,则不可能，若度为奇的等于二，则路线固定，其所经历所有点，若等于零则枚举起点

这里有个小技巧，a^a=0,0^a=a 所以异或时重复经历的点只要判经历次数是否为奇数即可

另外有个坑的是位运算优先级较低，你写

for(int i=1; i<=n; i++)
                {
                    if(ans^a[i]>mx)
                        mx=ans^a[i];
                }

就会wa，它先会判断>再异或

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define mod 10000007
#define maxn 100005

int a[maxn];
int p[maxn];
int pre[maxn];
int n,m,sz;

int fin(int x)
{
    if(x != pre[x])
        pre[x] = fin(pre[x]);
    return pre[x];
}

void unio(int x,int y)
{
    int x1=fin(x);
    int x2=fin(y);
    if(x1!=x2)
    {
        pre[x1]=x2;
        sz--;
    }
}

int main()
{
    int o,u,v;
    while(~scanf("%d",&o))
    {
        while(o--)
        {
            scanf("%d%d",&n,&m);
            for(int i=1; i<=n; i++)
            {
                scanf("%d",&a[i]);
            }
            memset(p,0,sizeof(p));
            for(int i=1; i<=n; i++)
            {
                pre[i]=i;
            }
            sz=n;

            for(int i=0; i<m; i++)
            {
                scanf("%d%d",&u,&v);
                unio(u,v);
                p[u]++;
                p[v]++;
            }

            if(sz>1)
            {
                printf("Impossible\n");
                continue;
            }

            int tot=0;
            for(int i=1; i<=n; i++)
            {
                if(p[i]%2==1)
                    tot++;
            }
            if(tot>2)
            {
                printf("Impossible\n");
            }
            else if(tot==2)
            {
                int ans=0;
                for(int i=1; i<=n; i++)
                {
                    if(((p[i]+1)/2)%2)
                        ans=ans^a[i];
                }
                printf("%d\n",ans);
            }
            else
            {
                int ans=0;
                for(int i=1; i<=n; i++)
                {
                    if(((p[i]+1)/2)%2) ans=ans^a[i];
                }
                int mx=0;
                for(int i=1; i<=n; i++)
                {
                    mx=max(mx,ans^a[i]);
                }
                printf("%d\n",mx);
            }
        }
    }
    return 0;
}