poj2299 查找逆序数 归并

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal 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

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05
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。}}<a target=_blank href="http://poj.org/searchproblem?field=source&key=Waterloo+local+2005.02.05" style="text-transform: none; text-decoration: none;"></a>