poj 1990 mooFest

I - MooFest

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit Status

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
x,y
对x按升序，y按升序排序之后
所以要求所有的费用，只需要考虑当前的cow和它前面每头cow的费用
要用到三个树状树状 
一个是记录当前cow前面有多少头cow的位置比它的位子小

一个是记录当前cow前面所有位置比它的位子小的总价值

一个是记录当前cow前面所有位置比它的位置大的总价值

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
#include<cstring>
#include<stdlib.h>
#include<string>
#include<cmath>
using namespace std;
#define pb push_back
int n;
__int64 p[20100],q[20100],qq[20100];
struct node{
    int v,x;
}so[20100];
int cmp(node a,node b){
    if(a.v==b.v) return a.x<b.x;
    return a.v<b.v;
}
void update1(int pos,int num){
    while(pos<=20000){
        p[pos]+=num;
        pos+=pos&(-pos);
    }
}
__int64 getnum1(int pos){
   __int64 sum=0;
    while(pos>=1){
        sum+=p[pos];
        pos-=pos&(-pos);
    }
    return sum;
}
void update2(int pos,int num){
    while(pos<=20000){
        q[pos]+=num;
        pos+=pos&(-pos);
    }
}
__int64 getnum2(int pos){
    __int64 sum=0;
    while(pos>=1){
        sum+=q[pos];
        pos-=pos&(-pos);
    }
    return sum;
}
void update3(int pos,int num){
    while(pos>=1){
        qq[pos]+=num;
        pos-=pos&(-pos);
    }
}
__int64 getnum3(int pos){
      __int64 sum=0;
        while(pos<=20000){
            sum+=qq[pos];
            pos+=pos&(-pos);
        }
        return sum;
}
int main(){
    while(cin>>n){
        memset(p,0,sizeof(p));
        memset(q,0,sizeof(q));
        for(int i=1;i<=n;i++)
            scanf("%d%d",&so[i].v,&so[i].x);
        sort(so+1,so+1+n,cmp);
        __int64 sum=0;
        for(int i=1;i<=n;i++){
            __int64 a,b,c;
            a=getnum1(so[i].x);//位置比当前位置小的cow的数量
            b=getnum2(so[i].x);//位置比当前位置小的总价值
            c=getnum3(so[i].x);//位置比当前位置大的总价值
            sum+=(a*so[i].x-b+c-(i-1-a)*so[i].x)*so[i].v;
            update1(so[i].x,1);
            update2(so[i].x,so[i].x);
            update3(so[i].x,so[i].x);
        }
        printf("%I64d
",sum);
    }
}