POJ1990 MooFest

　　POJ1990 MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3400		Accepted: 1330

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

*****************************************************************************

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：一开始水水的写了一个n^2的算法，这题终究没有那么白痴。原来是用了树状数组。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只和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。这里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。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define MAX(a,b) (a>b?a:b)
#define ABS(a) ((a)>0?(a):-1*(a))
#define TK(a) (a&(a^(a-1)))
#define N 20005
using namespace std;

struct Node
{
    long long v,x;
    bool operator<(const Node & a)const
    {
        return v<a.v;
    }
}node[N];

long long num[2][N];

long long rsum(int pos,int d)
{
    long long ans=0;
    while(pos>0)
    {
        ans+=num[d][pos];
        pos-=TK(pos);
    }
    return ans;
}

void toput(int pos,long long v,int d)
{
    while(pos<=20000)
    {
        num[d][pos]+=v;
        pos+=TK(pos);
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++)
        scanf("%lld%lld",&node[i].v,&node[i].x);
    sort(node+1,node+1+n);
    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        long long a=rsum(node[i].x,0),b=rsum(node[i].x,1);
        ans+=(node[i].x*a-b+rsum(20000,1)-b-(i-1-a)*node[i].x)*node[i].v;
        toput(node[i].x,1,0);
        toput(node[i].x,node[i].x,1);
    }
    printf("%lld\n",ans);
}