【树状数组】——poj1990——灵活运用

                                                 MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6551		Accepted: 2889

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

 

（转自——http://www.cnblogs.com/Fatedayt/archive/2011/10/08/2202439.html）

题目大意：一群牛参加完牛的节日后都有了不同程度的耳聋，

              第i头牛听见别人的讲话，别人的音量必须大于v[i]，当两头牛i，j交流的时候，交流的最小声音为max{v[i],v[j]}*他们之间的距离。

              现在有n头牛，求他们之间两两交流最少要的音量和。

 

解题思路：首先将这n头牛按照v值从小到大排序（后面说的排在谁的前面，都是基于这个排序）。这样，排在后面的牛和排在前面的牛讲话，两两之间所用的音量必定为后面的牛                 的v值，这样一来才有优化的余地。然后，对于某头牛i来说，只要关心跟排在他前面的牛交流就好了。我们必须快速地求出排在他前面的牛和他之间距离的绝对值之和                   ans，只要快速地求出ans，就大功告成。

              这里需要两个树状数组。树状数组可以用来快速地求出某个区间内和，利用这个性质，我们可以快速地求出对于牛i，x位置比i小牛的个数，以及这个牛的位置之和。

              这就需要两个树状数组，一个记录比x小的牛的个数a，一个记录比x小的牛的位置之和b，

              然后，我们可以快速地求出牛i和比牛i 位置小的牛的所有距离的绝对值为：a*x[i]-b;

              也可以方便地求出比牛i位置大的牛到牛i的距离和，即所有距离-b-(i-1-a)*x[i]；那么此题就差不多了。

      

//两个树状数组
#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;

#define maxn  200010

long long cow[maxn];//记录比i的能量小的牛的个数
long long dis[maxn];//记录比i小的牛的距离之和
struct ox{
    int eng;
    int pos;
}c[maxn];//记录每个牛的信息
#include<iostream>

bool cmp(ox a,ox b)//按照能量排序
{
    if(a.eng==b.eng)
        return a.pos<b.pos;
    return a.eng<b.eng;
}

long long sum(long long temp[maxn],int x)
{
    long long s = 0;
    while(x>0)
    {
      s+=temp[x];
      x-=(x&-x);
    }
    return s;
}

void add(long long temp[maxn],int x,int add)
{
    while(x<=maxn)
    {
       temp[x]+=add;
       x+=(x&-x);
    }
}

int main()
{
    long long ans=0;

    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>c[i].eng>>c[i].pos;
    }

    sort(c+1,c+n+1,cmp);

    memset(cow,0,sizeof(cow));
    memset(dis,0,sizeof(dis));


    for(int i=1;i<=n;i++)//计算每头牛的各种组合的和
    {
        long long total_dis = sum(dis,maxn-1);//所有距离总和
        long long total = sum(dis,c[i].pos);//到牛i的距离总和
        long long cnt = sum(cow,c[i].pos);//排在牛i前的牛的个数

        ans+= c[i].eng * (cnt*c[i].pos-total /*<-前面牛的距离；后面牛的距离->*/+ total_dis-total-(i-cnt-1)*c[i].pos);

        //更新
        add(dis,c[i].pos,c[i].pos);
        add(cow,c[i].pos,1);
    }

    printf("%lld\n",ans);

    return 0;
}