HDOJ 2689

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1940    Accepted Submission(s): 1390

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

3 1 2 3 4 4 3 2 1

 

 

Sample Output

0 6

 

 

Author

WhereIsHeroFrom

 

 

Source

ZJFC 2009-3 Programming Contest

 

 

Recommend

yifenfei

 

题意是给你一个序列，问你最少交换多少次可以使整个序列单调上升。

数据只有1000，直接n^2可做，但是有一种更为精妙的方法——树状数组 0Ms

我们先初始化整个序列为0，然后每输入一个数x，就在x位置将0换成1，此时求该位置的sum，就是比x小的数字数目，显然i - sum就是是这个数字要交换的次数。

#include<iostream>

#include<cmath>

#include<cstdio>

#include<cstring>

using namespace std;

int a[1010], c[1010] , n;

int lowbit(int x)

{

    return x & (-x);

}

int sum(int x)

{

    int sum = 0;

    while(x > 0)

    {

        sum += c[x];

        x -= lowbit(x);

    }

    return sum;

}

void update(int x , int num)

{

    while(x <= n)

    {

        c[x] += num;

        x += lowbit(x);

    }

}

int main()

{

    while(~scanf("%d" , &n))

    {

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

        memset(c , 0 , sizeof(c));

        int ans = 0;

        int x;

        for(int i = 1 ; i <= n ; i++)

        {

            scanf("%d" , &x);

            update(x , 1);

            ans += i - sum(x);

        }

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

    }

    return 0;

}