[AtCoder Beginner Contest 396] F - Rotated Inversions

题目描述

题目链接

解法一、树状数组求逆序对

#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long LL;

const int N = 2e5 + 10;
int n, m;
LL tr[N]; //tr[i]表示当前数组元素中值 <= i的有多少

int lowbit(int x) {
    return x & -x;
}

void add(int x, int c) {
    for(int i = x; i <= m; i += lowbit(i)) {
        tr[i] += c;
    }
}

LL sum(int x) {
    LL res = 0;
    for(int i = x; i > 0; i -= lowbit(i)) {
        res += tr[i];
    }
    return res;
}

int main() {
    cin >> n >> m;
    vector<int> a(n);
    
    for (int i = 0; i < n; i ++) {
        cin >> a[i];
    }
    
    //存储每个值在原数组中的位置
    vector<vector<int>> g(m);
    for (int i = 0; i < n; i ++) {
        g[a[i]].pb(i);
    }

    LL ans = 0;
    
    //利用树状数组进行数组初始时的逆序数(即k = 0时)
    //把值x存储到位置x + 1
    //那么对于当前位置为i的值a[i]，其位置在之前并比其值大的数组元素有sum(m) - sum(a[i] + 1)
    //然后在把当前的值的个数增加存入到树状数组中, 即add(a[i] + 1, 1)
    for(auto x : a) { // x = a[i]
        ans += sum(m) - sum(x + 1);
        add(x + 1, 1);
    }
    
    cout << ans << '\n';

    for (int c = 1; c < m; ++c) {
        LL c1 = 0, c2 = 0;
        for (int idx = 0; idx < g[m - c].size(); idx ++) {
            //如果x + c = m，在这一轮中x的值将会变为(x + c) % m = 0，逆序数就可能会发生变化
            //分别讨论增加与减少的数量，其值分别为c1、c2
            //在某个变为0的数之前有多少不为0的数呢？值为g[m - c][idx] - idx(减去了跟它同组变为0且在它前面的数)
            c1 += g[m - c][idx] - idx;
            //同理需要计算，某个变为0的数之后有多少不为0的数。值为(n - 1) - g[m - c][idx] - (g[m - c].size() - 1 - idx)
            c2 += n - 1 - g[m - c][idx] - (g[m - c].size() - 1 - idx);
        }
        
        ans += c1 - c2;
        cout << ans << '\n';
    }
    
    return 0;
}

解法二、归并排序求逆序对

#include<bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long LL;

const int N = 2e5 + 10;
int n, m;
LL tr[N]; //tr[i]表示当前数组元素中值 <= i的有多少

int lowbit(int x) {
    return x & -x;
}

void add(int x, int c) {
    for(int i = x; i <= m; i += lowbit(i)) {
        tr[i] += c;
    }
}

LL sum(int x) {
    LL res = 0;
    for(int i = x; i > 0; i -= lowbit(i)) {
        res += tr[i];
    }
    return res;
}

//求解数组arr的逆序数 
LL merge(vector<int>& arr, int left, int right) {
    if (left >= right) {
        return 0;
    }
    
    int mid = (left + right) >> 1;
    LL res = merge(arr, left, mid) + merge(arr, mid + 1, right);

    vector<int> temp(right - left + 1);
    int i = left, j = mid + 1, k = 0;
    
    while(i <= mid && j <= right) {
        if(arr[i] <= arr[j]) {
            temp[k ++] = arr[i ++];
        } else {
            res += mid - i + 1;
            temp[k ++] = arr[j ++];
        }
    }
    
    while(i <= mid) temp[k ++] = arr[i ++];
    while(j <= right) temp[k ++] = arr[j ++];
    for (int p = 0; p < k; p ++) {
        arr[left + p] = temp[p];
    }
    
    return res;
}

int main() {
    cin >> n >> m;
    vector<int> a(n);
    
    for (int i = 0; i < n; i ++) {
        cin >> a[i];
    }
    
    //存储每个值在原数组中的位置
    vector<vector<int>> g(m);
    for (int i = 0; i < n; i ++) {
        g[a[i]].pb(i);
    }

    LL ans = merge(a, 0, n - 1);
    cout << ans << '\n';

    for (int c = 1; c < m; ++c) {
        LL c1 = 0, c2 = 0;
        for (int idx = 0; idx < g[m - c].size(); idx ++) {
            //如果x + c = m，在这一轮中x的值将会变为(x + c) % m = 0，逆序数就可能会发生变化
            //分别讨论增加与减少的数量，其值分别为c1、c2
            //在某个变为0的数之前有多少不为0的数呢？值为g[m - c][idx] - idx(减去了跟它同组变为0且在它前面的数)
            c1 += g[m - c][idx] - idx;
            //同理需要计算，某个变为0的数之后有多少不为0的数。值为(n - 1) - g[m - c][idx] - (g[m - c].size() - 1 - idx)
            c2 += n - 1 - g[m - c][idx] - (g[m - c].size() - 1 - idx);
        }
        
        ans += c1 - c2;
        cout << ans << '\n';
    }
    
    return 0;
}