POJ 2299 Ultra-QuickSort 快速排序求逆序對

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 59356 Accepted: 21972
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

5
9
1
0
5
4
3
1
2
3
0
Sample Output

6
0

題目鏈接：http://poj.org/problem?id=2299

題意

給個數組，求逆序對的個數。

題解

求逆序對的個數是快速排序的一個應用，在快排過程中，當右半邊的數組復
制到t數組中時，左半邊剩下的元素個數m-p就是當前逆序對的個數,每次排
序加上這個m-p就是結果。

---

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <stack>
#include <queue>
#include <vector>
#define INF 0xffffffff
using namespace std;

const int maxn = 500000+5;
int a[maxn];
int t[maxn];

void merge_sort(int x, int y, long long int & ans){
    if(y-x>1){
        int m = x+(y-x)/2;
        int p=x, q=m, i=x;
        merge_sort(x, m, ans);
        merge_sort(m ,y, ans);
        while(p<m || q<y){
            if(q>=y || p<m && a[p]<=a[q]){
                t[i++] = a[p++];
            }
            else{
                t[i++] = a[q++];
                ans += m-p;
            }
        }
        for(int j=x; j<y; j++){
            a[j] = t[j];
        }
    }
}

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