Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 43816 | Accepted: 15979 |
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
题意：给出一个数列，每次交换相邻数字，求排成递增序的最少交换次数。
思路：本应该是冒泡排序求，但是数值为50w，所以铁定会超时。归并排序可以求序列的逆序数，序列的逆序数=在只允许相邻两个元素交换的条件下,得到有序序列的交换次数
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int input[500010];
int tmp[500010];
long long sum;
void merge(int l,int mid,int r)
{
int i=l;
int j=mid+1;
int k=1;
while(i<=mid&&j<=r){//这时候i 和 j 指向的部分都排序完毕了 现在合并
if(input[i]>input[j]){
tmp[k++]=input[j++];
sum+=mid-i+1;////第i个比j大 由于i已经从小到大排过序了 那么i+1到mid的也会比j大
}
else
tmp[k++]=input[i++];
}
while(i<=mid)
tmp[k++]=input[i++];
while(j<=r)
tmp[k++]=input[j++];
for(i=l,k=1;i<=r;i++,k++){
input[i]=tmp[k];
}
}
void merge_sort(int l,int r)
{
if(l<r){//长度大于1 这是个判断不是循环
int mid=(l+r)/2;
merge_sort(l,mid);
merge_sort(mid+1,r);
merge(l,mid,r);
}
}
int main()
{
int n,i;
while(~scanf("%d",&n)){
if(n==0)
break;
for(i=0;i<n;i++){
scanf("%d",&input[i]);
}
sum=0;
merge_sort(0,n-1);
printf("%lld\n",sum);
}
return 0;
}