HDU 5883 The Best Path (欧拉路或者欧拉回路)

时间：2023-03-09 23:42:56
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\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;

}

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