给定两个长度为n的排列a, b, 美丽值为|a1 - b1| + |a2 - b2| +...+|an - bn|，构造排列p，使得i -＞ p[i]，求最大美丽值-优快云博客

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本文介绍了一段C++代码，用于解决一个涉及路径遍历和哈希表的问题，通过计算路径中两个节点值对应的哈希值差的绝对值，求解给定数组中满足特定条件的路径总数。

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思路：

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e9 + 5, maxm = 4e4 + 5, mod = 1e9 + 7, N = 1e6;
int a[maxn], p[maxn], b[maxn], p2[maxn];
int n, m;
string s;
bool vis[maxn];

int qpow(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

void solve(){
    int res = 0;
    cin >> n;
    for(int i = 1; i <= n; i++){
        vis[i] = 0;
        p2[i] = 0;
        p[i] = 0;
    }
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    for(int i = 1; i <= n; i++){
        cin >> b[i];
        p[b[i]] = i;
    }
    int cnt = 0;
    int k = 1;
    vector<pair<int, int>> vec;
    for(int i = 1; i <= n; i++){
        vector<pair<int, int>> path;
        path.pb({a[i], b[i]});
        int j = i;
        int siz = 0;
        while(!vis[j]){
            vis[j] = 1;
            siz++;
            j = p[a[j]];
            path.pb({a[j], b[j]});
        }
        if(siz == 1){
            continue;
        }
        for(auto [x, y] : path){
            if(!p2[x] && !p2[y]){
                p2[x] = k;
                p2[y] = n - k + 1;
                res += 2LL * abs(p2[x] - p2[y]);
                k++;
            }
        }
    }

    cout << res << '\n';
}
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    // fac[0] = 1;
    // for(int i = 1; i <= N; i++){
    //     fac[i] = fac[i - 1] * i % mod;
    // }
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}