HDU 5883 The Best Path (欧拉路或者欧拉回路)

题意： $n$个点 m 条无向边的图，找一个欧拉通路/回路使得这个路径所有结点的异或值最大。

析：由欧拉路性质，奇度点数量为0或2。一个节点被进一次出一次，度减2，产生一次贡献，因此节点 i 的贡献为 i 点的度数除以2然后再模22​​degree​u​​​​⌋ mod 2)∗a​u​​。欧拉回路的起点贡献多一次，

欧拉通路的起点和终点贡献也多一次。因此如果是欧拉回路的话枚举一下起点就好了。

但是这个题有坑，就是有孤立点，这些点可以不连通，。。。。被坑死了，就是这一点，到最后也没过。。。伤心

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int p[maxn], in[maxn];
int Find(int x) {  return x == p[x] ? x : p[x] = Find(p[x]); }
int a[maxn];
int cnt;
vector<int> vv;

bool judge(){
    int x = Find(1);
    cnt = 0;vv.clear();
    for(int i = 1; i <= n; ++i){
        if(x != Find(i) && i != Find(i)) return false;
        if(in[i] & 1)  ++cnt, vv.push_back(i);
        if(cnt > 2)  return false;
    }

    if(cnt && cnt != 2)  return false;
    return true;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; ++i){
            p[i] = i;
            scanf("%d", &a[i]);
        }
        memset(in, 0, sizeof in);
        int u, v;
        for(int i = 0; i < m; ++i){
            scanf("%d %d", &u, &v);
            int x = Find(u);
            int y = Find(v);
            if(x != y)  p[y] = x;
            ++in[u];  ++in[v];
        }

        if(!m){ printf("0
");  continue; }
        if(!judge()){  printf("Impossible
");  continue;  }
        int ans = 0;
        for(int i = 1; i <= n; ++i){
            int t = in[i]/2;
            if(t & 1)  ans ^= a[i];
        }

        if(cnt){
            ans ^= a[vv[0]];
            ans ^= a[vv[1]];
        }
        else{
            int x = ans;
            for(int i = 1; i <= n; ++i){
                if(ans < (x ^ a[i])){
                    ans = x ^ a[i];
                }
            }
        }
        printf("%d
", ans);
    }
    return 0;
}