POJ 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 40008 Accepted: 14432

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

搜索思想 超时

#include <iostream>//由小到大排序using namespace std;int s[500005];int main(){    int i, j;    int n;    while(cin>>n&&n){for(i=1;i<=n;i++)cin>>s[i];int sum=0;int temp;for(i=1;i<n;i++)for(j=1;j<n;j++)if(s[j]>s[j+1]){sum=sum+1;temp=s[j];s[j]=s[j+1];s[j+1]=temp;}cout<<sum<<endl;}return 0;}

分冶思想——归并排序

#include <iostream>using namespace std;const int n=500000;int a[n],b[n];unsigned __int64 sum;void merge(int p,int q,int r){int begin_a=p,begin_b=q+1,t=p;for(;begin_a<=r;begin_a++)b[begin_a]=a[begin_a];begin_a=p;while((begin_a<=q) && (begin_b<=r)){if(b[begin_a]<=b[begin_b]){a[t++]=b[begin_a++];}else{a[t++]=b[begin_b++];sum+=q-begin_a+1;}}while(begin_a<=q)a[t++]=b[begin_a++];while(begin_b<=r)a[t++]=b[begin_b++];}void mergeSort(int low,int high){if(low<high){int mid=(low+high)/2;mergeSort(low,mid);mergeSort(mid+1,high);merge(low,mid,high);}}int main(){int i,m;while(scanf("%d",&m) && m!=0){for(i=0;i<m;i++)scanf("%d",&a[i]);sum=0;mergeSort(0,m-1);printf("%I64d\n",sum);}return 0;}