2299 Ultra-QuickSort

最新推荐文章于 2022-12-15 12:59:42 发布

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最新推荐文章于 2022-12-15 12:59:42 发布

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分类专栏： POJ 文章标签： input integer sorting merge output algorithm

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 24055		Accepted: 8603

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

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.

Output

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.

Sample Input

Sample Output

Source

Waterloo local 2005.02.05

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
#define MAX 500001

int n, a[MAX], t[MAX];
long long sum;

/* 归并 */
void Merge(int l, int m, int r) {
    /* p指向输出区间 */
    int p = 0;
    /* i、j指向2个输入区间 */
    int i = l, j = m + 1;
    /* 2个输入区间都不为空时 */
    while(i <= m && j <= r) {
        /* 取关键字小的记录转移至输出区间 */
        if (a[i] > a[j]) {
            t[p++] = a[j++];
            /* a[i]后面的数字对于a[j]都是逆序的 */
            sum += m - i + 1;
        }else {
            t[p++] = a[i++];
        }
    }
    /* 将非空的输入区间转移至输出区间 */
    while(i <= m) t[p++] = a[i++];
    while(j <= r) t[p++] = a[j++];
    /* 归并完成后将结果复制到原输入数组 */
    for (i = 0; i < p; i++){
        a[l + i] = t[i];
    }
}

/* 归并排序 */
void MergeSort(int l, int r) {
    int m;
    if (l < r) {
        /* 将长度为n的输入序列分成两个长度为n/2的子序列 */
        m = (l + r) / 2;
        /* 对两个子序列分别进行归并排序 */
        MergeSort(l, m);
        MergeSort(m + 1, r);
        /* 将2个排好的子序列合并成最终有序序列 */
        Merge(l, m, r);
    }
}

int main() {
    int i;
    while(cin >> n) {
        if (n == 0) break;
        sum=0;
        for(i = 0; i < n; i++) {
            scanf("%d", &a[i]);
        }
        MergeSort(0, n - 1);
        printf("%I64d\n", sum);
    }
    return 0;
}