POJ2299 Ultra-QuickSort

 

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 53685		Accepted: 19722

Description

题目太长可以不看。目的是给定数列求逆序对。

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

Source

Waterloo local 2005.02.05

 

归并排序求逆序对。

归并排序分分钟写完，但是longlong忘了用lld输出，又浪费了一阵子青春

 

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int mxn=520000;
long long a[mxn],t[mxn];
int n;
long long ans;
void msort(int l,int r){
    if(r-l>1)
    {
        int mid=l+(r-l)/2;
        msort(l,mid);
        msort(mid,r);
        int p=l,q=mid,i=l;//指向起点 
        while(p<mid || q<r){//范围内有数就继续处理 
            if(q>=r || (p<mid && a[p]<=a[q]))
            {
                t[i++]=a[p++];
            }
            else {t[i++]=a[q++];ans+=mid-p;};
        }
        for(i=l;i<r;i++)a[i]=t[i];//用排序后的序列覆盖原数组对应部分 
    }
    return;
}
int main(){
    while(scanf("%d",&n) && n){
        ans=0;
        int i,j;
        for(i=1;i<=n;i++)scanf("%lld",&a[i]);
        msort(1,n+1);
        printf("%lld\n",ans);
    }
    return 0;
}