POJ 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 52744		Accepted: 19331

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

求逆序数，和 http://blog.csdn.net/sdfgdbvc/article/details/51171769 这题一模一样，可以复习一下

//归并排序
#include <stdio.h>
#define MAX 500005
int N;
int a[MAX], b[MAX];
long long sum;
void Merge(int start, int mid, int end)
{
int i = start, j = mid + 1, k = start;
while(i <= mid && j <= end)
{
if(a[i] <= a[j])
{
b[k++] = a[i++];
}
else
{
sum += j - k;
b[k++] = a[j++];
}
}
while(i <= mid)
{
b[k++] = a[i++];
}
while(j <= end)
{
b[k++] = a[j++];
}
for(i = start; i <= end; i++)
a[i] = b[i];
}
void MergeSort(int start, int end)
{
if(end > start)
{
int mid = (start + end) / 2;
MergeSort(start, mid);
MergeSort(mid+1, end);
Merge(start, mid, end);
}
}
int main()
{
int i;
while(scanf("%d", &N) && N)
{
sum = 0;
for(i = 0; i < N; i++)
scanf("%d", &a[i]);
MergeSort(0, N-1);
printf("%lld\n", sum);
}
return 0;
}
/*
//树状数组
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 500005
int N;
int tree[MAX], f[MAX];
struct P
{
int v;
int num;
} pro[MAX];
int cmp(const void *a, const void *b)
{
return (*(P*)a).v - (*(P*)b).v;
}
int lowbit(int x)
{
return x & -x;
}
void updata(int n, int x)
{
while(n <= N)
{
tree
+= x;
n += lowbit(n);
}
}
int find(int n)
{
int sum = 0;
while(n >= 1)
{
sum += tree
;
n -= lowbit(n);
}
return sum;
}
int main()
{
int i;
long long sum;
while(scanf("%d", &N) != EOF && N)
{
sum = 0;
memset(tree, 0, sizeof(tree));
for(i = 1; i <= N; i++)
{
scanf("%d", &pro[i].v);
pro[i].num = i;
}
qsort(pro+1, N, sizeof(pro[1]), cmp);
int id = 1;
f[pro[1].num] = 1;
for(i = 2; i <= N; i++)
{
if(pro[i].v == pro[i-1].v)
f[pro[i].num] = id;
else
f[pro[i].num] = ++id;
}
for(i = 1; i <= N; i++)
{
updata(f[i], 1);
sum += i - find(f[i]);
}
printf("%lld\n", sum);
}
return 0;
}*/