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

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-;

const int N=+,maxn=+,inf=0x3f3f3f3f;

int s[][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=;

    while(i>){

        ans+=s[d][i];

        i-=i&(-i);

    }

    return ans;

}

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

 //   cout<<setiosflags(ios::fixed)<<setprecision(2);

    int n;

    while(cin>>n){

        memset(s,,sizeof s);

        for(int i=;i<=n;i++)cin>>e[i].v>>e[i].x;

        sort(e+,e++n,comp);

        ll ans=;

        for(int i=;i<=n;i++)

        {

            ll a=sum(e[i].x,),b=sum(e[i].x,);

        //    cout<<sum(N,1)-b-(i-1-a)*e[i].x<<endl;

            ans+=(a*e[i].x-b+sum(N,)-b-(i--a)*e[i].x)*e[i].v;

            add(e[i].x,,);//0是比x小的牛的个数

            add(e[i].x,e[i].x,);//1是比x小的牛的距离和

        }

        cout<<ans<<endl;

    }

    return ;

}

秒客网

poj1990树状数组

相关文章