Ultra-QuickSort （归并排序，逆序数）

Posted 2020-12-24 albert-biu

题目描述

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

 

输入

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

 

输出

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

 

样例输入

5
9
1
0
5
4
3
1
2
3
0

 

样例输出

6
0

#include <bits/stdc++.h>

using namespace std;

const int maxn = 5e5 + 1000;
const int inf = 0x3f3f3f3f;

int s[maxn], t[maxn];
long long cnt = 0;
int n;

void merge(int l1, int r1, int l2, int r2) {
    int k = 0;
    while (l1 <= r1 && l2 <= r2) {
        if (s[l2] < s[l1]) t[++k] = s[l2++], cnt += r1 - l1 + 1;
            //若s[l2]小，则l1~r1都比a[j]要大，他们都会与a[j]构成逆序对
        else t[++k] = s[l1++];
    }
    while (l1 <= r1) t[++k] = s[l1++];
    while (l2 <= r2) t[++k] = s[l2++];
    for (int i = k, j = r2; i >= 1; i--, j--) {//j的最终值不一定为1~r2
        s[j] = t[i];
    }
}

void mergesort(int l, int r) {
    int mid = (l + r) / 2;
    if (l < r) {
        mergesort(l, mid);
        mergesort(mid + 1, r);
        merge(l, mid, mid + 1, r);
    }
}

int main() {
    freopen("input.txt", "r", stdin);
    while (cin >> n) {
        cnt = 0;
        if (n == 0)
            break;
        for (int i = 1; i <= n; i++)
            cin >> s[i];
        mergesort(1, n);
        cout << cnt << endl;
    }
    return 0;
}