归并排序求逆序数
Time Limit:7000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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
题目大意：
给出长度为n的序列，每次只能交换相邻的两个元素，问至少要交换几次才使得该序列为递增序列。
刚刚学了时间复杂度， 用归并排序Mergesort了，O(nlogn)，省时，不会超时。
这里用归并排序并不是为了求交换次数，而是为了求序列的逆序数，而一个乱序序列的逆序数 = 在只允许相邻两个元素交换的条件下,得到有序序列的交换次数。
案例中的
9 1 0 5 4
要把它排列为升序0，1，4，5，9
而对于序列9 1 0 5 4
9后面却有4个比9小的元素，因此9的逆序数为4
1后面只有1个比1小的元素0，因此1的逆序数为1
0后面不存在比他小的元素，因此0的逆序数为0
5后面存在1个比他小的元素4, 因此5的逆序数为1
4是序列的最后元素，逆序数为0
因此序列9 1 0 5 4的逆序数 t=4+1+0+1+0 = 6 ，就是交换次数
注意：保存逆序数时，必须要用long long型定义，会WA的。。。
#include<iostream>
using namespace std;
long long total;
int n,a[500005];
int t[500005];
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++];
total+=m-p;
}
}
for(i=x; i<y; i++)
a[i]=t[i];
}
}
int main()
{
while(cin>>n&&n)
{
total=0;
for(int i=0; i<n; i++)
cin>>a[i];
merge_sort(a,0,n,t);
cout<<total<<endl;
}
return 0;
}