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=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 inline void write(int x)
 {
     if(x<)
     {
         putchar('-');
         x=-x;
     }
     if(x>)
     {
         write(x/);
     }
     putchar(x%+'');
 }
 const int N=;
 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=;
     while(x>)
     {
         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,,sizeof(c));
         memset(num,,sizeof(num));
         for(int i=;i<=n;i++)
         {
             scanf("%d",&q[i].x);
             q[i].id=i;
         }
         sort(q+,q++n,cmp);
         for(int i=;i<=n;i++)
         {
             num[q[i].id]=i;
         }
         int sum = ;
         for(int i=;i<=n;i++)
         {
             add(num[i],);
             sum+=getsum(n)-getsum(num[i]);
         }
         printf("%d\n",sum);
     }
     return ;
 }