poj 2299

[align=center]Ultra-QuickSort[/align]

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48121		Accepted: 17553

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
题意：
给出n个数，要求求出按照冒泡排序法需要多少次才能把这n个数按从小到大的顺序排列。
思路：
需要多少次才能按从小到大的顺序排列，可以转化成求其逆序数（这个线代学过），但是n<=500000,逆序数可能会超过int ,所以结果要用long long ，否则wrong answer.求其逆序数可以用归并排序，按照分治三步:
1.划分问题：吧序列分成元素个数尽量相等的两半。
2.递归求解：把元素分别排列。
3.把两个有序表合并成一个。
中间过程需要引入一个辅助空间T
.
每一次累加cnt+=m-p;即是答案。
代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
int  A[500004],T[500004];
long long  cnt;
using namespace std;
void init(int n)
{
for(int i=0;i<n;i++)
scanf("%d",&A[i]);

}
void merge_sort(int *A,int x,int y,int *T)
{
if(y-x>1)
{
int m=x+(y-x)/2;
int p=x,q=m,i=x;
merge_sort(A,x,m,T);
merge_sort(A,m,y,T);
while(p<m||q<y)
{
if(q>=y||(p<m&&A[p]<=A[q]))    T[i++]=A[p++];
else
{
T[i++]=A[q++];
cnt+=m-p;

}

}            for(i=x;i<y;i++)
A[i]=T[i];
}
}
int main()
{
int n;
while(scanf("%d",&n)&&n)
{
init(n);
cnt=0;
memset(T,0,sizeof(T));
merge_sort(A,0,n,T);//不能把n写成n-1，否则会答案错误
printf("%I64d\n",cnt);
}

return 0;
}