HDu 2686 Sort is (树状数组）

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#树状数组

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本文介绍了一种通过计算逆序数来确定将无序序列排序所需的最少交换次数的方法。利用树状数组进行高效计算，提供了完整的代码实现，并解释了核心算法原理。

Sort it

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4744 Accepted Submission(s): 3301

Problem Description

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.

Output

For each case, output the minimum times need to sort it in ascending order on a single line.

Sample Input

Sample Output

0
6
他们有的用冒泡排序水过这道题，这道题的正解是用树状数组，然后求逆序数。
树状数组入门博客：http://blog.youkuaiyun.com/passer__/article/details/76653144
题意:通过多少次可以将给的数字变成从小到大的序列。

#include<stdio.h>//逆序数 
#include<string.h>
int a[1050];//刷新数字 
int c[1050];//储存有当前这个数字前边有多少比他大的的数字 
int  lowbit(int x)//树状 
{
    return x&(-x);
}
void sum1(int x,int y)
{
    int sum=0;
    while(x>0)
    {
        sum=sum+a[x];
        x=x-lowbit(x);
    }
    c[y]=y-1-sum;//除去自己 用已经放进去的数字减去他前边有多少比他小的数字 就是前边有多少比它大的数字
}
void sum(int x) 
{
    while(x<=1010)
    {
        a[x]++;
        x=x+lowbit(x);
    }
} 
int main()
{
    int n,x;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        { 
            scanf("%d",&x);
            sum1(x,i);//查找当前数字前有多少比他小的
            sum(x);
        }
        int sum=0;
        for(int i=1;i<=n;i++)
            sum=sum+c[i];
        printf("%d\n",sum);    
    }
    return 0;
}