HDU 2689 Sort it【树状数组】

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4672    Accepted Submission(s): 3244

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2689

分析：好吧，我智障了，听说有种很快的写法，但好像跑的速度没我快？应该是我代码跑的最快吧，写法也是最复杂的，这题我是用树状数组乱搞的，反正15ms，相当快啊！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
inline void write(int x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
const int N=1001;
int n;
int num[N];
int c[N];
struct node
{
    int x;
    int id;
}q[N] ;
bool cmp(node a,node b)
{
    return a.x<b.x;
}
int lowbit(int x)
{
    return x&(-x);
}
int getsum(int x)
{
    int s=0;
    while(x>0)
    {
        s+=c[x];
        x-=lowbit(x);
    }
    return s;
}
void add(int x,int y)
{
    while(x<=n)
    {
        c[x]+=y;
        x+=lowbit(x);
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(c,0,sizeof(c));
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&q[i].x);
            q[i].id=i;
        }
        sort(q+1,q+1+n,cmp);
        for(int i=1;i<=n;i++)
        {
            num[q[i].id]=i;
        }
        int sum = 0;
        for(int i=1;i<=n;i++)
        {
            add(num[i],1);
            sum+=getsum(n)-getsum(num[i]);
        }
        printf("%d
",sum);
    }
    return 0;
}