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最新推荐文章于 2020-08-11 08:53:56 发布

HyJoker

最新推荐文章于 2020-08-11 08:53:56 发布

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分类专栏：树状数组

本文链接：https://blog.youkuaiyun.com/HyJoker/article/details/52153140

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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

Sample Output

  
  
   
   7


   
   
    
    
     
     Hint
    
    
Input Details

Three cows are standing in line with respective grumpiness levels 2, 3, and 1.
Output Details

2 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).

题意：这题其实还挺好理解的，就是看每个数t之前有多少个比他大的数（设为x），然后用比他大的数的和sum加上x*t，所以很明显要用树状数组来做。

纠结了好久的树状数组怎么更新和维护，看了别人的博客之后大概懂了一点。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<vector>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
long long ans[100010];
long long Count[100010];
long long n;
void join(long long k,long long num){
	while(k<=n){
		ans[k]+=num;
		Count[k]+=1;
		k+=k&-k;
	}
}
long long Sum(long long k){
	long long sum=0;
	while(k){
		sum+=ans[k];
		k-=k&-k;
	}
	return sum;
}
long long Sum_count(long long k){
	long long sum=0;
	while(k){
		sum+=Count[k];
		k-=k&-k;
	}
	return sum;
}
int main(){
#ifdef LOCAL
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	memset(ans,0,sizeof(ans));
	while(scanf("%d",&n)!=EOF){
		long long res=0;
		for(int i=1;i<=n;i++){
			long long temp;
			scanf("%lld",&temp);
			join(temp,temp);
			long long k1=i-Sum_count(temp);//到此为止  比x大的个数；  
			///sum_count[x] 为输入i个数的时候 x之前有sum_count[x]个比x小的数 用i相减则为大于x的个数
			if(k1!=0){
				long long k2=Sum(n)-Sum(temp);//Sum(n)-Sum(temp)为到当前为止比这个数大的所有数之和。
				res+=k1*temp+k2;
			}
		}
		printf("%lld\n",res);
	}
    return 0;
}