poj 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 44601		Accepted: 16214

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

 

分析：

求逆序对，因此利用归并排序，在排序过程中，记录逆序对的数目。

什么是逆序对

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 #define max 500000
 5 int c[max+5];
 6 int a[max+5];
 7 long long count;
 8 void Merge(int *a,int first,int last){
 9     int p,q,temp,mid;
10     mid=(first+last)/2;
11     temp=first;
12     p=first;
13     q=mid+1;
14     while(p<=mid&&q<=last){
15         if(a[p]>a[q]){
16             count+=mid+1-p;
17             c[temp++]=a[q++];
18         }
19         else{
20             c[temp++]=a[p++];    
21         }
22     }
23     while(q<=last){ 
24             c[temp++]=a[q++];
25     }
26     q--;
27     while(p<=mid){
28             //count+=q-mid;
29             c[temp++]=a[p++];
30     }
31     int i=first;
32     for(;i<=last;i++){
33         a[i]=c[i];
34     }
35 }
36 void MergeSort(int *a,int first,int last){
37     if(first==last){
38         return;
39     }
40     int mid=(first+last)/2;
41     MergeSort(a,first,mid);
42     MergeSort(a,mid+1,last);
43     Merge(a,first,last);
44 }
45 int main(){
46     int n;
47     while(scanf("%d",&n)&&n){
48         count=0;
49         int i=0;
50         for(;i<n;i++){
51              scanf("%d",&a[i]);
52         }
53         MergeSort(a,0,n-1);
54          cout<<count<<endl;
55     }
56     //cout<<1<<endl;
57     return 0;
58 }

 

转载于:https://www.cnblogs.com/Deribs4/p/4279953.html