1poj2299(串，归并排序)

http://poj.org/problem?id=2299

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 27554		Accepted: 9860

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：求数组元素的逆序数个数。直接用暴力(冒泡排序)tle..网上有很好的解答
http://www.slyar.com/blog/poj-2299-c.html
归并排序http://zh.wikipedia.org/wiki/归并排序
代码
//代码1：tle

#include<stdio.h>

#include<string.h>

int a[500010];

int main()

{

int n;

while(scanf("%d",&n)!=EOF)

{

int i,j;

if(n==0)

break;

int count=0;

memset(a,0,sizeof(a));

for(i=0;i<n;i++)

scanf("%d",&a[i]);

for(i=0;i<n;i++)

{

for(j=0;j<i;j++)

{

if(a[j]>a[i])

{

int
t=a[i];

a[i]=a[j];

a[j]=t;

count++;

}

}

}

printf("%d\n",count);

}

return 0;

}

代码2：AC

#include<stdio.h>

#define MAXN 500001

int n,a[MAXN],t[MAXN];

__int64 sum;

void Merge(int l,int m,int r)

{

int p=0;// p指向输出区间

int i=l,j=m+1;// i、j指向2个输入区间

while(i<=m&&j<=r)

{

if(a[i]>a[j])//前面一个元素比后面一个元素大满足逆序sum累加.取关键字小的记录转移至输出区间

{

t[p++]=a[j++];//
a[i]后面的数字对于a[j]都是逆序的

sum+=m-i+1;

}

else

t[p++]=a[i++];

}

while(i<=m)t[p++]=a[i++];//
将非空的输入区间转移至输出区间

while(j<=r)t[p++]=a[j++];

for(i=0;i<p;i++)//
归并完成后将结果复制到原输入数组

a[l+i]=t[i];

}

void MergeSort(int l,int r)

{

int m;

if(l<r)

{

m=(l+r)/2;//
将长度为n的输入序列分成两个长度为n/2的子序列

MergeSort(l,m);//
对两个子序列分别进行归并排序

MergeSort(m+1,r);

Merge(l,m,r);//将2个排好的子序列合并成最终有序序列

}

}

int main()

{

int i;

while(scanf("%d",&n)!=EOF)

{

if(n==0)

break;

sum=0;

for(i=0;i<n;i++)

{

scanf("%d",&a[i]);

}

MergeSort(0,n-1);//从0到最后一个元素n-1

printf("%I64d\n",sum);

}

return 0;

}