hdu 2838（树状数组求逆序数）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2838

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2382    Accepted Submission(s): 766

Problem Description

Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.

Please help Sherlock calculate the minimal time required to reorder the cows.

 

Input

Line 1: A single integer: N
Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.

 

Output

Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.

 

Sample Input

3231

 

Sample Output

7
Hint
Input DetailsThree cows are standing in line with respective grumpiness levels 2, 3, and 1.Output Details2 3 1 : Initial order.2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4).1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).
 

 

Source

2009 Multi-University Training Contest 3 - Host by WHU 

思路： eg:  4 5 3 2 1  我们可以观察到第三个数 的rank值是3，那么在这个数的左边有 两个比它大的数，说明第三个数至少得和左边的数交换2次 ~~

           so ~    设一个数a左边有k个比他大的数  并且这k个数的和是sum，  那么交换代价是 cnt=k*a+sum；对吧~所以遍历一遍即可，（用树状数组加速，可以用来求逆序数）

    

#include <iostream>#include <stdio.h>#include <string.h>#include <cstdio>#include <cmath>typedef long long ll;const int N=1e5+100;using namespace std;ll c[N];  ll index[N]; 

ll a[N];int n;int lowbit(int x){    return x&(-x);}void update(int x,int d){  while(x<=n)  {    c[x]+=d;    index[x]+=1;    x+=lowbit(x);  }}ll get_sum(int x){ ll ans=0; while(x>0) {   ans+=c[x];   x-=lowbit(x); } return ans;}ll get_index(int x){  ll ans=0;  while(x>0)  {    ans+=index[x];    x-=lowbit(x);  }  return ans;}void Init(){  for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);  memset(c,0,sizeof(c));  memset(index,0,sizeof(index));}int main(){        while(scanf("%d",&n)!=EOF)        {          Init();          ll cnt=0;          for(int i=1;i<=n;i++)          {            update(a[i],a[i]);            ll k=i-get_index(a[i]);            if(k)            {             cnt+=k*a[i];             cnt+=get_sum(n)-get_sum(a[i]);            }          }          printf("%I64d\n",cnt);        }        return 0;}