ive namespace accepted return ali Once bmi wan spec

The Best Path

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2104 Accepted Submission(s): 841

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

解析 先判斷圖是否只有一個連通塊 然後判斷是否存在歐拉回路或者路徑 歐拉路徑每個點經過 (度數+1)/2 次 歐拉回路也一樣 但是起點多走一次 每個點都可以作為起點

所以直接枚舉取最大值就好了。

AC代碼

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e5+10;
 4 int a[maxn],du[maxn],par[maxn];
 5 int _find(int x)
 6 {
 7     return x==par[x]?x:par[x]=_find(par[x]);
 8 }
 9 void unio(int a,int b)
10 {
11     int ra=_find(a);
12     int rb=_find(b);
13     if(ra!=rb) par[rb]=ra;
14 }
15 int main()
16 {
17     int t;
18     scanf("%d",&t);
19     while(t--)
20     {
21         int n,m;
22         scanf("%d%d",&n,&m);
23         for(int i=1;i<=n;i++)
24         {
25             par[i]=i;du[i]=0;
26             scanf("%d",&a[i]);
27         }
28         for(int i=0;i<m;i++)
29         {
30             int u,v;
31             scanf("%d%d",&u,&v);
32             unio(u,v);
33             du[u]++,du[v]++;
34         }
35         int sum=0,flag=0;
36         for(int i=1;i<=n;i++)
37         {
38             if(du[i]&1)sum++;
39             if(i!=1&&par[i]!=par[i-1])flag=1;
40         }
41         if(sum>2||flag)
42             printf("Impossible\n");
43         else
44         {
45             int ans=0;
46             for(int i=1;i<=n;i++)
47             {
48                 int temp=(du[i]+1)/2;
49                 while(temp)
50                     ans=ans^a[i],temp--;
51             }
52             if(sum==0)
53             {
54                 int temp=ans;
55                 for(int i=1;i<=n;i++)
56                     ans=max(temp^a[i],ans);
57             }
58             printf("%d\n",ans);
59         }
60     }
61 }

HDU 5883 歐拉路徑異或值最大 水題