逆序数 poj2299 Ultra-QuickSort

Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 59643 Accepted: 22101
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
题意：这个题的意思是问你冒泡排序中交换值的次数，也就是让你求序列中的逆数对之和；

求在数列的逆序数可以使用归并排序。归并排序是将数列a[l,h]分成两半a[l,mid]和a[mid+1,h]分别进行归并排序，然后再将这两半合并起来。在合并的过程中（设l<=i<=mid，mid+1<=j<=h），当a[i]<=a[j]时，并不产生逆序数；当a[i]>a[j]时，在前半部分中比a[i]大的数都比a[j]大，将a[j]放在a[i]前面的话，逆序数要加上mid+1-i。因此，可以在归并排序中的合并过程中计算逆序数

这个代码好像只能在pku上ac；那就粘贴第二份代码，这是重做之后发现的错误

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int n;int A[500005],T[500005];__int64 ans;void mersort(int x,int y){    if(y-x<=1)        return;    int mid=(x+y)/2;    mersort(x,mid);    mersort(mid,y);    int p=x,i=x,q=mid;    while(p<mid||q<y)    {        if(q==y||(p<mid&&A[p]<=A[q]))            T[i++]=A[p++];        else if(p==mid||(A[q]<A[p]))        {            if(p<mid) ans+=(mid-p);            T[i++]=A[q++];        }    }    for(int i=x; i<y; i++)    {        A[i]=T[i];    }}int main(){    while(~scanf("%d",&n)&&n)    {        memset(A,0,sizeof(A));        memset(T,0,sizeof(T));        for(int i=0; i<n; i++)        {            scanf("%d",&A[i]);        }        ans=0;        mersort(0,n);        printf("%I64d\n",ans);    }    return 0;}

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define LL long long intint a[500005],b[500005];int n;LL ans;//极限情况500000*500000会爆，所以开LLvoid mersort(int x,int y){    if(y-x<=1)        return ;    int mid=(x+y)/2;    mersort(x,mid);    mersort(mid,y);    int p=x,i=x,q=mid;    while(p<mid||q<y)    {        if(q==y||(p<mid&&a[p]<=a[q]))//注意，不是a[p]<=a[mid],弄清比较的对象        //x到y排序，所以应该和a[q]比，从而把后面的一到前面去。            b[i++]=a[p++];        else if(p==mid||a[p]>a[q])        {            if(p<mid) ans+=(mid-p);            b[i++]=a[q++];        }    }    for(int i=x; i<y; i++)        a[i]=b[i];}int main(){    while(~scanf("%d",&n)&&n)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        ans=0;        mersort(0,n);        printf("%lld\n",ans);    }    return 0;}