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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\n"); continue; }
if(!judge()){ printf("Impossible\n"); 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\n", ans);
}
return 0;
}
HDU 5883 The Best Path (欧拉路或者欧拉回路)
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5879545.html
