POJ 1990 MooFest 树状数组

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9189		Accepted: 4158

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

Sample Output

Source

USACO 2004 U S Open

题意给你若干牛的声贝值和他们的x坐标叫你求每两个牛之间的距离乘上他们两个的最大声贝值

我们怎么分析呢？我们这样考虑

大声贝的牛出现会影响乘法的声贝值

那么我们按照声贝排序按小到大的顺序这样每次我乘法的声贝值都是取新加的这个牛

这个牛x坐标前面有几个牛呢？维护一个lsum 左边的牛那么右边的牛数目自然为i-1-lsum

那么左边的牛的距离和是多少呢？是lsum * x - lsumdisx(左边牛的距离和) 那么右边牛的距离自然为sumdisx(总距离和)-lsumdisx-(i-1-lsum)*x

那么我们这样树状数组维护一个数量前缀数组a

距离前缀和数组b 就可以实现了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX_N = 20024;
long long a[MAX_N],b[MAX_N],xsum[MAX_N];
struct node {
    int x;
    int v;
    bool operator < ( const node &t) const {
         return v<t.v;
    }
}col[MAX_N];
long long getsum(int x,long long *arr){
    long long ans = 0;
    for(int i = x;i;i-=i&(-i)) ans+=arr[i];
    return ans;
}
void change(int x,long long dis,long long *arr,long long v){
   for(;x<=dis;x+=x&(-x)) arr[x]+=v;
}

int main(){
    int n;
    scanf("%d",&n);
    long long dis = -1;
    for(int i=1;i<=n;++i){
        scanf("%d%d",&col[i].v,&col[i].x);
        if(col[i].x>=dis) dis= col[i].x;
    }
    sort(col+1,col+1+n);
    long long ans = 0;
    for(int i=1;i<=n;++i) xsum[i] = xsum[i-1]+col[i].x;
    for(int i=1;i<=n;++i){
      long long dl = getsum(col[i].x,a);
      long long dr = i-1-dl;
      long long lsum = getsum(col[i].x,b);
      long long rsum = xsum[i-1]-lsum;
      ans+=col[i].v*(dl*col[i].x-lsum);
      ans+=col[i].v*(rsum-dr*col[i].x);
      change(col[i].x,dis,a,1);
      change(col[i].x,dis,b,col[i].x);
    }
    printf("%lld\n",ans);
    return 0;
}

POJ 1990 MooFest 树状数组

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