The Best Path

最新推荐文章于 2020-11-25 20:40:43 发布

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最新推荐文章于 2020-11-25 20:40:43 发布

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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

Sample Output

  
  
   
   2
Impossible
  
  




/*欧拉路的判断，
寻找最多的幸运数;
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int big = 1e5 + 7;
int a[big];//每个节点的度数;
int  b[big];//湖的编号;
int z[big], e, f, flag;
bool judge(int n)//判断是否为欧啦路;
{
    int sum = 0, h;
    for(int i = 1; i <= n; i ++)
        if(z[i] == i) sum ++;
    if(sum != 1) return false;
    sum = flag = 0;
    for(int i = 1; i <= n; i ++)
    {
        if(a[i]&1)
        {
            sum ++;
            a[i] ++;
        }
        h = a[i]/2;
        if(!(h&1)) b[i] = 0;
        }
    if(sum != 0 && sum != 2) return false;
    if(sum == 2) flag = 1;
    return true;
}
void G(int n)//湖的标号;
{
    for(int i = 1; i <= n; i ++)
        z[i] = i;
}
int Find(int key)
{
    int s = key;
    while(s != z[s])
        s = z[s];
    int j = key, e;
    while(j != s)
    {
        e = z[j];
        z[j] = s;
        j = e;
    }
    return s;
}
int main()
{
    int t, n, m, u, v, p, q;
    scanf("%d", &t);
    while(t --)
    {
        scanf("%d %d", &n, &m);
        memset( a, 0, sizeof( a));
        for(int i = 1; i <= n; i ++)
            scanf("%d", &b[i]);
        G(n);//湖的编号;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d %d", &u, &v);
            a[u] ++;
            a[v] ++;
            p = Find(u);
            q = Find(v);
            if(p == q) continue;
            if(p > q) swap( p, q);
            z[q] = p;
        }
        if(judge(n))
        {
            int temp = b[1], ans = 0;
            for(int i = 2; i <= n; i ++)
                temp = temp xor b[i];
            if(flag) ans = temp;
            else
            {
                for(int i = 1; i <= n; i ++)
                    ans = max( ans, temp xor b[i]);
            }
            printf("%d\n", ans);
        }
        else printf("Impossible\n");
    }
    return 0;
}
#include<bits/stdc++.h>
using namespace std;
const int maxn = 112345;
int arr[maxn];
int fnd(int x){
    return x == arr[x] ? x : arr[x] = fnd(arr[x]);
}
void join(int x,int y){
    x = fnd(x),y = fnd(y);
    if(x != y)
        arr[x] = y;
}
int val[maxn],ind[maxn];
int cal(int n){
    int bio = 0,odd = 0;
    for(int i=1;i<=n;i++){
        if(arr[i] == i && ind[i] != 0)
            bio++;
        odd += ind[i] % 2;
    }
    if(bio >= 2)
        return -1;
    if(bio == 0){
        return *max_element(val+1,val+1+n);
    }
    if(odd == 0){
        int all = 0,ret = 0;
        for(int i=1;i<=n;i++){
            if((ind[i]/2) & 1) all ^= val[i];
        }
        for(int i=1;i<=n;i++){
            if(ind[i])
                ret = max(ret,all ^ val[i]);
        }
        return ret;
    }
    if(odd == 2){
        int ret = 0;
        for(int i=1;i<=n;i++){
            if(ind[i]){
                if(ind[i] % 2){
                    if(((ind[i]+1)/2)&1) ret ^= val[i];
                }
                else{
                    if((ind[i]/2)&1) ret ^= val[i];
                }
            }
        }
        return ret;
    }
    return -1;
}


int main(){
    int T;
    scanf("%d",&T);
    int n,m;
    while(T-- && ~scanf("%d %d",&n,&m)){
        memset(ind,0,sizeof(ind));
        for(int i=1;i<=n;i++){
            scanf("%d",&val[i]);
            arr[i] = i;
        }
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d %d",&x,&y);
            join(x,y);
            ind[x]++,ind[y]++;
        }
        int cnt = cal(n);
        if(cnt == -1){
            puts("Impossible");
        }
        else{
            printf("%d\n",cnt);
        }
    }
    return 0;
}