poj1990树状数组

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
我觉得我需要收回当初说树状数组比线段树简单这句话。。太坑了，一题比一题坑。。完全按题解写的
题意：给v【i】，x【i】要求所以的牛的音量和即x【i】-x【j】*max（v【i】，v【j】）之和
题解：两个树状数组数组一起使用，一个求x之前的比x坐标小的数(a)，一个求x之前的比x坐标小的坐标和(b);
那么比x小的坐标和x的坐标的总坐标差是a*(e[i].x)-b;比x大的坐标和x的坐标的总坐标差是总坐标-b-(i-1-a)*e[i].x

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-9;
const int N=20000+5,maxn=32000+5,inf=0x3f3f3f3f;

int s[2][N];
struct edge{
    ll v,x;
}e[N];

bool comp(const edge &a,const edge &b)
{
    return a.v<b.v;
}
void add(int i,ll x,int d)
{
    while(i<=N){
        s[d][i]+=x;
        i+=i&(-i);
    }
}
ll sum(int i,int d)
{
    ll ans=0;
    while(i>0){
        ans+=s[d][i];
        i-=i&(-i);
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int n;
    while(cin>>n){
        memset(s,0,sizeof s);
        for(int i=1;i<=n;i++)cin>>e[i].v>>e[i].x;
        sort(e+1,e+1+n,comp);
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            ll a=sum(e[i].x,0),b=sum(e[i].x,1);
        //    cout<<sum(N,1)-b-(i-1-a)*e[i].x<<endl;
            ans+=(a*e[i].x-b+sum(N,1)-b-(i-1-a)*e[i].x)*e[i].v;
            add(e[i].x,1,0);//0是比x小的牛的个数
            add(e[i].x,e[i].x,1);//1是比x小的牛的距离和
        }
        cout<<ans<<endl;
    }
    return 0;
}