树状数组（好）-poj1990

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MooFest

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4549 Accepted: 1895

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

43 12 52 64 3

Sample Output

57现在这智商太着急了。。。首先将这n头牛按照v值从小到大排序。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]下面是代码：
#include<iostream>#include<set>#include<map>#include<vector>#include<queue>#include<cmath>#include<climits>#include<cstdio>#include<string>#include<cstring>#include<algorithm>typedef long long LL;using namespace std;const int MAX=20005;struct node{    LL v,x;    friend bool operator<(node a,node b)    {        return a.v<b.v;    }}cow[MAX];int N;LL num[2][MAX];int low(int x){    return x&(-x);}LL sum(int x,int biao){    LL ans=0;    while(x>0)    {        ans+=num[biao][x];        x-=low(x);    }    return ans;}void update(int x,LL v,int biao){    while(x<=20000)    {        num[biao][x]+=v;        x+=low(x);    }}int main(){    //freopen("in.txt","r",stdin);    while(cin>>N)    {        for(int i=1;i<=N;i++)            scanf("%lld%lld",&cow[i].v,&cow[i].x);        sort(cow+1,cow+N+1);        memset(num,0,sizeof(num));        LL ans=0;        for(int i=1;i<=N;i++)        {            LL a=sum(cow[i].x,0),b=sum(cow[i].x,1);            ans+=(a*cow[i].x-b+sum(20000,1)-b-(i-1-a)*cow[i].x)*cow[i].v;            update(cow[i].x,1,0);            update(cow[i].x,cow[i].x,1);        }        cout<<ans<<endl;    }    return 0;}