hdoj5883The Best Path【歐拉迴路】_hdoj5883The

The Best Path Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 233    Accepted Submission(s): 102


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  Nlines 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 2016 ACM/ICPC Asia Regional Qingdao Online  

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<map>#include<set>using namespace std;typedef long long LL;typedef pair<int,int>pii;const int maxn=1000010;int num[maxn];int degree[maxn];int father[maxn];int Find(int x){    return x==father[x]?x:father[x]=Find(father[x]);}int main(){    int t,n,m;cin>>t;    while(t--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;++i){            scanf("%d",&num[i]);            father[i]=i;degree[i]=0;        }int ans=0;        for(int i=1;i<=m;++i){            int a,b;            scanf("%d%d",&a,&b);            degree[a]++;degree[b]++;            a=Find(a);b=Find(b);            if(a!=b)father[a]=b;        }        bool sign=true;        int root=Find(1);        for(int i=1;i<=n;++i){            if(Find(i)!=root){                sign=false;break;            }        }        int cnt=0;        for(int i=1;i<=n;++i){            if(degree[i]&1)cnt++;        }        if(sign&&(cnt==0||cnt==2)){            ans=0;            if(cnt==2){                for(int i=1;i<=n;++i){                    if(degree[i]&1){                        ans^=num[i];                        if(((degree[i]-1)/2)&1){                            ans^=num[i];                        }                    }                    else {                        if(((degree[i])/2)&1){                            ans^=num[i];                        }                    }                }                printf("%d\n",ans);            }            else {                int cntx=0;                for(int i=1;i<=n;++i){                    if((degree[i]/2)&1){                        cntx^=num[i];                    }                }                for(int i=1;i<=n;++i){                    ans=max(ans,cntx^num[i]);                }                printf("%d\n",ans);            }        }        else {            printf("Impossible\n");        }    }    return 0;}