MooFest_二维树状数组

Description

Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing. 

Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)). 

Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume. 

Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location. 

Output

* Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows. 

Sample Input

4
3 1
2 5
2 6
4 3

Sample Output

57

【题意】有n头牛，排列成一条直线，给出每头牛在直线上的坐标d。每头牛有一个v，如果牛i和牛j想要沟通的话，它们必须用max(v[i],v[j])，消耗的能量为：max(v[i],v[j]) * 它们之间的距离.

问要使所有的牛之间都能沟通(两两之间)，总共需要消耗多少能量。

【思路】现将v从小到大排列，使得每次取到的是当前最大的v。

c[1]记录当前牛的数量c[2]记录当前所有牛的d之和。（二维树状数组）

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=20000;
struct node
{
    int d,v;
    bool operator <(const node &a)const
    //从小到大排序，使得当前获得的v一定是出现过最大的。
    {
        return v<a.v;
    }
}moo[N+10];
int c[3][N+10];
int lowbit(int x)
{
    return x&(-x);
}
void update(int i,int d,int v)
{
    while(d<=N)
    {
        c[i][d]+=v;
        d+=lowbit(d);
    }
}
int get_sum(int i,int d)
{
    int res=0;
    while(d)
    {
        res+=c[i][d];
        d-=lowbit(d);
    }
    return res;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            scanf("%d%d",&moo[i].v,&moo[i].d);
        sort(moo+1,moo+1+n);
        int sum=0;//记录所有坐标之和
        long long int ans=0;
        for(int i=1;i<=n;i++)
        {
            int d=moo[i].d;
            sum+=d;
            update(1,d,1);//c[1]记录牛数量
            update(2,d,d);//c[2]记录牛坐标之和
            int n1=get_sum(1,d);//在i牛及他前面有多少头
            int n2=get_sum(2,d);//在i牛及他前面的牛坐标和为多少
            int tmp1=n1*d-n2;//i左边的坐标差
            int tmp2=sum-n2-d*(i-n1);//i右边的坐标差
            ans+=(long long int)(tmp1+tmp2)*moo[i].v;
           //不用longlong会溢出
        }
        printf("%lld
",ans);
    }
    return 0;
}