POJ 2299 Ultra-QuickSort(归并排序·逆序对)

标签：acm   归并排序   

题意  给你一个数组求其中逆序对(i<j&&a[i]>a[j])的个数 

我们来看一个归并排序的过程:
给定的数组为[2, 4, 5, 3, 1]，二分后的数组分别为[2, 4, 5], [1, 3]，假设我们已经完成了子过程，现在进行到该数组的“并”操作：

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 500005;
int a[N], t[N], n;
long long cnt;

void merge(int l, int m, int r)
{
    int pl = l, pr = m + 1, p = 0;
    while(pl <= m && pr <= r)
    {
        if(a[pl] <= a[pr]) t[p++] = a[pl++];
        else
        {
            t[p++] = a[pr++];
            cnt += m + 1 - pl;
        }
    }
    while(pl<=m) t[p++] = a[pl++];
    while(pr<=r) t[p++] = a[pr++];
    memcpy(a + l, t, sizeof(int)*p);
}

void mergeSort(int l, int r)
{
    if(l >= r) return;
    int	m = (l + r) >> 1;
    mergeSort(l, m);
    mergeSort(m + 1, r);
    merge(l, m, r);
}

int main()
{
    while(scanf("%d", &n), n)
    {
        for(int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        cnt = 0;
        mergeSort(0, n - 1);
        printf("%lld\n", cnt);
    }
    return 0;
}

Description

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.

Input

Output

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

POJ 2299 Ultra-QuickSort(归并排序·逆序对)

标签：acm   归并排序   

原文地址：http://blog.csdn.net/acvay/article/details/44984309