1015（树状数组+数学运算）

Problem 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 <br> <br>* 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. <br>

Output

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

Sample Input

4
3 1
2 5
2 6
4 3

Sample Output

57

题目大概：

两个点之间有一个权值为他们之间最大的音量值，他们之间的价值是距离差乘权值。求所有牛之间的总价值。

思路：

先按照权值不同，升序排序。

两个树状数组，一个算比他小的牛的位置之和，一个计算比他位置小的牛的数量。

然后从权值高的开始计算，价值=比他位置小的点的距离之差*权值+比他位置大的距离之差*权值

最后都加起来。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

long long c[20010][2];

struct poin
{
    long long l,v;
}a[20010];
int cmp(const poin a,const poin b)
{
    return a.v<b.v;
}
long long lowbit(long long x)
{
    return x&(-x);
}
int add(long long x,long long v,int k)
{
    while(x<=20001)
    {
        c[x][k]+=v;
        x=x+lowbit(x);
    }
}

long long sum(long long x,int k)
{
    long long su=0;
    while(x>0)
    {
        su+=c[x][k];
        x-=lowbit(x);
    }
    return su;
}
int main()
{
    int t;

    while(~scanf("%d",&t))
    {


    memset(c,0,sizeof(c));
    long long ss=0;
    for(int i=1;i<=t;i++)
   {

        long long w1,w2;
        scanf("%I64d%I64d",&w1,&w2);
        ss+=w2;
        a[i].v=w1;a[i].l=w2;
        add(w2,1,0);
        add(w2,w2,1);
    }
    sort(a+1,a+t+1,cmp);
    long long sun=0;
    for(long long i=t;i>0;i--)
    {

        long long e1=sum(a[i].l-1,0);
        long long e2=sum(a[i].l-1,1);
        long long sum1,sum2;
        long long w1,w2;
        w1=t-e1-1; w2=ss-e2-a[i].l;
        sum1=(e1*a[i].l-e2)*a[i].v;
        sum2=(w2-w1*a[i].l)*a[i].v;
        sun+=sum1+sum2;
        add(a[i].l,-1,0);
        add(a[i].l,-a[i].l,1);
        t--;
        ss=ss-a[i].l;
    }

    printf("%I64d\n",sun);

    }
    return 0;
}