uva 10810 - Ultra-QuickSort（归并求逆序数）

Problem B: Ultra-QuickSort

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.

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.

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

Output for Sample Input

6
0

Stefan B�ttcher

归并法求逆序数。关键思想，在归并过程中，如果用n^2的做法求左右两部分间的逆序数是不会减少复杂度的。但如果左右两部分已经是有序的，就可以用 two pointers 的方法，用O(n)合并。那对左右两边先排序会影响最终答案吗？很明显不会。。。归并排序恰好可以在归并求解的过程同时完成，所以用归并排序更快。我试了直接用sort排序，也能通过，慢一些，但写起来简单。

#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 500000 + 5;
const int INF = 1000000000;
typedef long long LL;
typedef pair<LL, int> P;

int a[maxn];
int tem[maxn];

LL solve(int l, int r){
    if(l>=r) return 0;
    LL ret = 0;
    int mid = (l+r)/2;
    ret += solve(l, mid);
    ret += solve(mid+1, r);

    int pl = l, pr = mid+1;
    for(int pl = l;pl <= mid;pl++){
        for(int j = pr;j <= r;j++){
            if(a[pl]<a[j]){
                break;
            }
            else
                pr++;
        }
        ret += pr-mid-1;
    }
/*
    pl = l, pr = mid+1;
    int cnt = l;
    while(pl <= mid || pr <= r){
        if(pl > mid){
            tem[cnt++] = a[pr++];
        }
        else if(pr > r){
            tem[cnt++] = a[pl++];
        }
        else if(a[pl]<a[pr]){
            tem[cnt++] = a[pl++];
        }
        else{
            tem[cnt++] = a[pr++];
        }
    }
    for(int i = l;i <= r;i++){
        a[i] = tem[i];
    }
*/
    sort(a+l,a+r+1);
    return ret;
}

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