poj 2299 Ultra-QuickSort 初级->数据结构->排序-归并排序

                                             
Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K

Total Submissions: 30020 Accepted: 10729

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： 求逆序数。

思想：利用归并排序求。

// /* poj 2299 利用归并排序求逆序数
// 3776 KB 1282 ms
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[500001],b[500001];
long long sum,n;
void mersort(int L,int R)
{
if(L>=R) return ;
int mid=L+((R-L)>>1);
mersort(L,mid);
mersort(mid+1,R);
int i=L,k=L,j=mid+1;
while(i<=mid&&j<=R)
{
if(a[i]>a[j])
{
b[k++]=a[j++];
sum+=mid-i+1;
}
else b[k++]=a[i++];
}
while(i<=mid) b[k++]=a[i++];
while(j<=R) b[k++]=a[j++];
for(i=L;i<=R;i++)
a[i]=b[i];
}
int main()
{
int i;
while(cin>>n,n)
{
for(i=1;i<=n;i++)
cin>>a[i];
sum=0;
mersort(1,n);
cout<<sum<<endl;
}
return 0;
}
// */

// /* 速度提升3倍多的求逆序数的归并算法，
// 只求结果，不排序，所以速度快。牛逼
// 3764 KB 375 ms
#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
const int N = 500010;

int num
, t
;
long long n;

long long merge_sort(int l, int r)
{
int mid, p, q, i, j, len;
long long ans = 0;

if(l >= r)    return 0;
mid = (l + r) >> 1;
len = r - l + 1;

ans += merge_sort(l, mid);
ans += merge_sort(mid + 1, r);

p = l; q = mid + 1;
j = l;

for(i = 0; i < len; ++i)
if((q>r)|| (num[p]<num[q]&&p<=mid))
t[j++] = num[p++];
else
{
ans += mid - p + 1;
t[j++] = num[q++];
}
for(i = l; i <= r; ++i)
num[i] = t[i];
return ans;
}
int main()
{
int i;
while(cin>>n,n)
{
for(i = 0; i < n; ++i)
scanf("%d", num + i);
printf("%lld\n",merge_sort(0,n-1));
for(i=1;i<=n;i++)
cout<<num[i];
}
return 0;
}
// */