poj 1990 MooFest (树状数组)

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4888		Accepted: 2058

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

Source

USACO 2004 U S Open

//600K    63MS    C++    1220B    2014-05-04 20:52:18
/* 

    题意：
        知道v和x，求每对间 max(vi,vj)与abs(vi-vj)的乘积的和，对v排序，然后每个v的计算公式：
            ans+=p[i].v*(cnt*p[i].x-sum+tsum-sum-(i-cnt-1)*p[i].x);
        知道计算的方法还是挺简单的！ 

*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define N 20005
struct node{
    int v,x;
}p[N];
__int64 cnum[N],clen[N];
int cmp(const void*a,const void*b)
{
    return (*(node*)a).v-(*(node*)b).v;
}
int lowbit(int k)
{
    return (-k)&k;
}
void update(__int64 c[],int k,int detal)
{
    for(;k<N;k+=lowbit(k))
        c[k]+=detal;
}
__int64 getsum(__int64 c[],int k)
{
    __int64 s=0;
    for(;k>0;k-=lowbit(k))
        s+=c[k];
    return s;
}
int main(void)
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(cnum,0,sizeof(cnum));
        memset(clen,0,sizeof(clen));
        for(int i=1;i<=n;i++)
            scanf("%d%d",&p[i].v,&p[i].x);
        qsort(p+1,n,sizeof(p[0]),cmp);
        __int64 ans=0;
        update(cnum,p[1].x,1);
        update(clen,p[1].x,p[1].x);
        for(int i=2;i<=n;i++){
            __int64 cnt=getsum(cnum,p[i].x);
            __int64 sum=getsum(clen,p[i].x);
            __int64 tsum=getsum(clen,N-1);
            ans+=p[i].v*(cnt*p[i].x-sum+tsum-sum-(i-cnt-1)*p[i].x);
            update(cnum,p[i].x,1);
            update(clen,p[i].x,p[i].x);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}