TOJ 2452 Ultra-QuickSort

描述

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.
输入

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.

输出

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.

样例输入

5
9
1
0
5
4
3
1
2
3
0

样例输出

6
0

题目来源

Waterloo Feb.5 2005

求最小的交换次数，即求逆序对数问题，并归排序。

【并归排序】

如下演示并归排序的整个过程：

并归排序主要是深搜实现的。

{9,1,0,4,5,8,7,4,3}=>{9,1,0,4,5} {8,7,4,3}

{9,1,0,4,5}=>{9,1,0} {4,5}=>{9}{1}{0} {4}{5}

{8,7,4,3}=>{8,7} {4,3}=>{8}{7} {4}{3}

合并子表
{0,1,9} {4,5} {7,8} {3,4}

{0,1,4,5,9} {3,4,7,8}

{0,1,3,4,4,7,8,9}

#include <stdio.h>
__int64 sum;
void mergeSort(int* a,int low,int mid,int high){
int* p=new int[high+1];
int i=low;
int j=low;//左侧表的起始位置
int h=mid+1;//右侧表的起始位置
while(h<=high&&j<=mid){
if(a[j]<=a[h]){
p[i]=a[j];
j++;
i++;
}else{
//求逆序数
sum+=h-i;
p[i]=a[h];
h++;
i++;
}
}
for(;j<=mid;j++,i++){
p[i]=a[j];
}
for(;h<=high;h++,i++){
p[i]=a[h];
}
for(i=low;i<=high;i++){
a[i]=p[i];
}
delete[] p;
}

void merge(int* a,int low,int high){
if(low<high){
int mid=(low+high)>>1;
//划分子表
merge(a,low,mid);
merge(a,mid+1,high);
//合并子表
mergeSort(a,low,mid,high);
}
}

int main()
{
int n;
int arr[500010];
while(scanf("%d",&n)!=EOF && n){
for(int i=0; i<n; i++){
scanf("%d",&arr[i]);
}
sum=0;
merge(arr,0,n-1);
printf("%I64d\n",sum);
}
return 0;
}