POJ 1990 MooFest

Posted 2020-12-25 bzmd

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

4
3 1
2 5
2 6
4 3

Sample Output

57

Source

USACO 2004 U S Open

 

首先简要说明题意：

给定2个序列 a，b

输出所有的i,j(i!=j）对应abs(a[j]-a[i])*max(b[i],b[j])

 

首先枚举所有的n方算法非常好想但是时间复杂度过大

我们思考这个式子应该如何提取公因式

abs(a[j]-a[i])这个部分每对i,j都不同

而max(b[i],b[j]),我们可以考虑对于最大的数，所有与它构成的数对都是乘这个b，然后对于次大的数时，我们就可以不考虑最大的数，其它的数乘这个b，。。。

如何快速计算各个abs（a[j]-a[i]）的和呢？

我们可以把它分成两部分

a[j]>a[i]  sum(a[j])-a[i]*sl1;

a[i]<a[j] a[i]*sl2-sum(a[j])

首先我们把所有数根据a排序，把它的a值放入树状数组中那么就可以的到sum(a[j])分不同的区间就可以的到比它大和比它小的坐标和

再根据b的大小排序，从大往小扫，还有我们要把这个数对应的位置变为0，这样才能够做到考虑后面的数不会算入

还有对于sl1和sl2,我们可以另外建一个树状数组先把每个位置都标为1，然后每做完一个就更改为0，查询的时也像上面一样。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=20005;
struct A
{
    int z,wz,id;
}a[N];
int c[N][2],n;
bool cmp1(const A &t1,const A &t2)
{
    return t1.wz<t2.wz;
}
bool cmp2(const A &t1,const A &t2)
{
    return t1.z<t2.z;
}
void add(int x,int h,int xh)
{
    for(;x<=n;x+=x&-x)
        c[x][xh]+=h;
}
int askk(int x,int xh)
{
    int re=0;
    for(;x;x-=x&-x)
        re+=c[x][xh];
    return re;    
}
int main()
{
    long long ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%d%d",&a[i].z,&a[i].wz),add(i,1,1);
    sort(a+1,a+n+1,cmp1);
    for(int i=1;i<=n;++i)
        a[i].id=i,add(i,a[i].wz,0);
    sort(a+1,a+n+1,cmp2);
    for(int i=n;i;--i)
    {
        add(a[i].id,-1,1); 
        add(a[i].id,-a[i].wz,0);
        ans+=(long long)a[i].z*(askk(n,0)-2*askk(a[i].id,0)+a[i].wz*(askk(a[i].id,1)*2-askk(n,1)));
    }
    printf("%lld",ans);
    return 0;    
}