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MooFest_二维树状数组

2024-08-15 18:15:40   221人阅读

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

 

 

 

【题意】有n头牛，排列成一条直线，给出每头牛在直线上的坐标d。每头牛有一个v，如果牛i和牛j想要沟通的话，它们必须用max(v[i],v[j])，消耗的能量为：max(v[i],v[j]) * 它们之间的距离.

问要使所有的牛之间都能沟通(两两之间)，总共需要消耗多少能量。

【思路】现将v从小到大排列，使得每次取到的是当前最大的v。

c[1]记录当前牛的数量c[2]记录当前所有牛的d之和。（二维树状数组）

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int N=20000;struct node{    int d,v;    bool operator <(const node &a)const    //从小到大排序，使得当前获得的v一定是出现过最大的。    {        return v<a.v;    }}moo[N+10];int c[3][N+10];int lowbit(int x){    return x&(-x);}void update(int i,int d,int v){    while(d<=N)    {        c[i][d]+=v;        d+=lowbit(d);    }}int get_sum(int i,int d){    int res=0;    while(d)    {        res+=c[i][d];        d-=lowbit(d);    }    return res;}int main(){    int n;    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++)            scanf("%d%d",&moo[i].v,&moo[i].d);        sort(moo+1,moo+1+n);        int sum=0;//记录所有坐标之和        long long int ans=0;        for(int i=1;i<=n;i++)        {            int d=moo[i].d;            sum+=d;            update(1,d,1);//c[1]记录牛数量            update(2,d,d);//c[2]记录牛坐标之和            int n1=get_sum(1,d);//在i牛及他前面有多少头            int n2=get_sum(2,d);//在i牛及他前面的牛坐标和为多少            int tmp1=n1*d-n2;//i左边的坐标差            int tmp2=sum-n2-d*(i-n1);//i右边的坐标差            ans+=(long long int)(tmp1+tmp2)*moo[i].v;           //不用longlong会溢出        }        printf("%lld\n",ans);    }    return 0;}

 

MooFest_二维树状数组


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