The Best Path
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 852 Accepted Submission(s): 359
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 aP0 XOR aP1 XOR ... XOR aPt. 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.
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
Source
解析：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 100000+5;
int a[MAXN];
int degree[MAXN];
int n, m;
void solve()
{
memset(degree, 0, sizeof(degree));
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
int u, v;
while(m--){
scanf("%d%d", &u, &v);
++degree[u];
++degree[v];
}
int odd_sum = 0;
for(int i = 1; i <= n; ++i){
if(degree[i]&1)
++odd_sum;
}
if(!(odd_sum == 0 || odd_sum == 2)){
printf("Impossible
");
return;
}
int val = 0;
for(int i = 1; i <= n; ++i){
if((degree[i]/2)&1)
val ^= a[i];
}
if(odd_sum == 0){
int res = 0xffffffff;
for(int i = 1; i <= n; ++i){
if(degree[i] != 0)
res = max(res, val^a[i]);
}
printf("%d
", res);
}
else{
int s = 0, e = 0;
for(int i = 1; i <= n; ++i){
if(degree[i]&1){
if(s == 0){
s = i;
}
else{
e = i;
break;
}
}
}
int res = val^a[s]^a[e];
printf("%d
", res);
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--){
solve();
}
return 0;
}