2016 ACM/ICPC Asia Regional Qingdao Online HDU5883

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5883

解法：先判断是不是欧拉路，然后枚举

#pragma comment(linker, "/STACK:102400000,102400000")
#include <math.h>
#include <time.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <iostream>
#include <algorithm>
#define pb push_back
#define fi first
#define se second
#define icc(x) (1<<(x))
#define lcc(x) (1ll<<(x))
#define lowbit(x) (x&-x)
#define debug(x) cout<<#x<<"="<<x<<endl
#define rep(i,s,t) for(int i=s;i<t;++i)
#define per(i,s,t) for(int i=t-1;i>=s;--i)
#define mset(g, x) memset(g, x, sizeof(g))
using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef double db;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> veci;
const int mod=(int)1e9+7,inf=0x3fffffff,rx[]={-1,0,1,0},ry[]={0,1,0,-1};
const ll INF=1ll<<60;
const db pi=acos(-1),eps=1e-8;template<class T> void rd(T &res){
    res = 0; int ch,sign=0;
    while( (ch=getchar())!='-' && !(ch>='0'&&ch<='9'));
    if(ch == '-') sign = 1; else res = ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res = (res<<3)+(res<<1)+ch-'0';
    res = sign?-res:res;
}
template<class T>void rec_pt(T x){
    if(!x)return;
    rec_pt(x/10);
    putchar(x%10^48);
}
template<class T>void pt(T x){
    if(x<0) putchar('-'),x=-x;
    if(!x)putchar('0');
    else rec_pt(x);
}
template<class T>inline void ptn(T x){ pt(x),putchar('\n'); }
template<class T>inline void Max(T &a,T b){ if(b>a)a=b; }
template<class T>inline void Min(T &a,T b){ if(b<a)a=b; }
template<class T>inline T mgcd(T b,T d){ return b?mgcd(d%b,b):d; }//gcd模板，传入的参数必须是用一类型
//-------------------------------主代码--------------------------------------//int g[100100];
int d[100100];
int markpath[100100];
int mark[100100];struct node{
    int to,next;
}edge[1000100];int cnt,pre[100100];void add_edge(int u,int v){
    edge[cnt].to = v;
    edge[cnt].next = pre[u];
    pre[u] = cnt++;
}void dfs(int s){
    mark[s] = 1;
    for(int p=pre[s];p!=-1;p=edge[p].next){
        int v = edge[p].to;
        if(mark[v] == 1) continue;
        dfs(v);
    }
}int main()
{
    int T;
    rd(T);
    while (T--) {
        cnt = 0;
        mset(d, 0); mset(markpath, 0); mset(pre, -1); mset(mark, 0);
        int n,m;
        rd(n),rd(m);
        for(int i=1;i<=n;i++)
        {
            rd(g[i]);
        }
        rep(i, 0, m){
            int x,y;
            rd(x),rd(y);
            markpath[x] = 1; markpath[y] = 1;
            add_edge(x,y);
            add_edge(y,x);
            d[x]++; d[y]++;
        }
        int cnt = 0;
        for(int i=1;i<=n;i++){
            if(d[i]%2 != 0) cnt++;
        }        if(cnt!=0 && cnt!=2){
            printf("Impossible\n");
            continue;
        }
        int flag = 0;
        for(int i=1;i<=n;i++){
            if(markpath[i]==1 && mark[i] ==0){
                flag ++;
                dfs(i);
            }
        }
        if(flag > 1) {
            printf("Impossible\n");
            continue;
        }        int ans = 0;        if(cnt == 0){
            int sum = 0;
            for(int i=1;i<=n;i++){
                if( (d[i]/2)%2!=0 )
                {
                    sum ^= g[i];
                }
            }
            for(int i=1;i<=n;i++){
                if(markpath[i]==1)
                    ans = max(ans,sum^g[i]);
            }
        }else{
            int sum = 0;
            for(int i=1;i<=n;i++){
                if(d[i]%2!=0){
                    d[i]++;
                }
                if( (d[i]/2)%2!=0 )
                {
                    sum ^= g[i];
                }
            }
            ans = sum;
        }
        ptn(ans);
    }
    return 0;
}