poj2299 查找逆序数 归并
Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 49032 | Accepted: 17936 |
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
Source
Waterloo local 2005.02.05
查找逆序数
查找逆序数
/* 由于本题实际上就是要求逆序对（即满足i<j，a[i]>a[j]的数对）的个数。而我们再回顾一下归并排序的过程： 假设回溯到某一步，后面的两部分已经排好序（就是说当前需要归并的两个部分都是分别有序的），假设这两个序列为 序列a1：2 3 5 9 序列a2：1 4 6 8 此时我们的目的就是要将a1和a2合并为一个序列。 由于在没排序前a2序列一定全部都是在a1序列之后的，当我们比较a2的1与a1的2时，发现1<2按照归并的思想就会先记录下a2的1，而这里实际上就是对冒泡排序的优化，冒泡是将a2的1依次与a1的9,5,3,2交换就需要4次，而归并却只有一次就完成了，要怎么去记录这个4呢，实际上由于1比2小而2后面还有4个数，也就是说那我的结果就必须要+4，也就是记录a1序列找到第一个比a2某一个大的数，他后面还余下的数的个数就是要交换的次数。 同时我们看a2的4时，a1中第一个比它大的数是5,5之后共有两个数，那结果就+2,。依次下去就可以计算出结果。但是由于我们任然没有改变归并排序的过程。所以复杂度还是O(nlogn) 此题有一坑就是结果会超int32,，用__int64*/ /*
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int a[500050],Arr[500050];
long long int Sort(int low,int mid,int high)
{
int i=low,j=mid+1,k=low;
long long ans=0;
while(i<=mid&&j<=high)
{
if(a[i]<=a[j])
Arr[k++]=a[i++];
else
{
Arr[k++]=a[j++];
ans+=j-k;
}
}
while(i<=mid)
{
Arr[k++]=a[i++];
}
while(j<=high)
{
Arr[k++]=a[j++];
}
for(int i=low;i<=high;i++)
{
a[i]=Arr[i];
}
return ans;
}
long long int MergeSort(int low,int high)
{
if(low>=high) //勿漏
return 0;
long long num=0;
int mid=(low+high)/2;
num+=MergeSort(low,mid);
num+=MergeSort(mid+1,high);
num+=Sort(low,mid,high);
return num;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n==0)
break;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
printf("%I64d\n",MergeSort(0,n-1)); //只有64位才不WA。
// for(int i=0;i<n;i++)
// cout<<a[i];
}
}
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int a[500050],Arr[500050];
long long num=0;
long long int Sort(int low,int mid,int high)
{
int i=low,j=mid+1;
int k=low;
while(i<=mid&&j<=high)
{
if(a[i]<=a[j]) //
Arr[k++]=a[i++];
else
{
Arr[k++]=a[j++];
num+=j-k; //逆序数的统计
}
}
while(i<=mid)
{
Arr[k++]=a[i++];
}
while(j<=high)
{
Arr[k++]=a[j++];
}
for(int i=low;i<=high;i++)
{
a[i]=Arr[i];
}
}
long long int MergeSort(int low,int high)
{
if(low>=high) //勿漏
return 0;
int mid=(low+high)/2;
MergeSort(low,mid);
MergeSort(mid+1,high);
Sort(low,mid,high);
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(n==0)
break;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
num=0;
MergeSort(0,n-1);
printf("%I64d\n",num);//只有64位才不WA。
}
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