參考自:點選打開連結
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
解題思路:
這個題就不能像之前那樣直接求逆序對了,一定會逾時(o(n^2))。
采用歸并算法的原理,複雜度降到o(nlog(n)),具體分析可百度,cnt=mid-i+1。
#include<stdio.h>
# define N 500050
int a[N];
long long int cnt;
void merge(int l,int mid,int r)
{
int i=l,j=mid+1,k=l;//
int tmp[N];
while(i<=mid&&j<=r)
{
if(a[i]<=a[j])
{
tmp[k++]=a[i++];
}
else
{
tmp[k++]=a[j++];
cnt+=mid-i+1;//求逆序對
}
}
while(i<=mid)
tmp[k++]=a[i++];
while(j<=r)
tmp[k++]=a[j++];
for(i=l;i<=r;i++)//
a[i]=tmp[i];
}
void merge_sort(int l,int r)
{
if(l<r)
{
int mid=(l+r)/2;
merge_sort(l,mid);
merge_sort(mid+1,r);
merge(l,mid,r);
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF&&n)
{
int i;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
cnt=0;
merge_sort(1,n);
printf("%lld\n",cnt);
}
return 0;
}