Poj 1990 -- moofest (2 tree arrays)

Moofest

Time limit:1000 ms		Memory limit:30000 K
Total submissions:5235		Accepted:2260

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 special 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

--------------------Split line--------------

Question:

Given the hearing and coordinates of an nheaded ox, it is required for every two cows to talk (max (V (I), V (j) * ABS (DIS [I]-Dis [J]). A total of N * (n-1)/2 (n/1) conversations are required.

Ideas:

Sort the listening of the ox from small to large, so the I-th ox conversation needs to be calculated.

1: Number A, coordinate, and B of cattle whose coordinates are smaller than the number I of cattle

2: Number of cows whose coordinates are greater than the number of cows whose heads are I. C, coordinates, and D

In this case, we need:

(D-C * A [I]. Y + A * A [I]. Y-B) * A [I]. x

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#define M 20000+10#define ll __int64using namespace std;ll b[M],c[M];struct node{    int x,y;    bool operator<(const node&a)const    {        return x<a.x;    }}a[M];void update(int x,int v,ll *h){    for(int i=x;i<=M;i+=i&-i){        h[i]+=v;    }}ll getsum(int x,ll *h){    ll sum=0;    for(int i=x;i>0;i-=i&-i){        sum+=h[i];    }    return sum;}int main(){    int n;    scanf("%d",&n);    memset(b,0,sizeof(b));    memset(c,0,sizeof(c));    for(int i=1;i<=n;++i){        scanf("%d %d",&a[i].x,&a[i].y);    }    sort(a+1,a+1+n);    ll ans=0;    for(int i=1;i<=n;++i){        ll x1=(ll)(getsum(a[i].y,b)*a[i].y-getsum(a[i].y,c));        ll x2=(ll)(getsum(M,c)-getsum(a[i].y,c)-(i-1-getsum(a[i].y,b))*a[i].y);        ans+=(ll)(x1+x2)*a[i].x;        update(a[i].y,1,b);        update(a[i].y,a[i].y,c);    }    printf("%I64d\n",ans);    return 0;}

Poj 1990 -- moofest (2 tree arrays)