树状数组 MooFest

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
  题解：
  这道题想了挺久的，主要一直没搞懂这个点与后面点的距离，只专注于树状数组的录入应从小到大。
  然后今天豁然开朗了。首先，用结构体，按音量从小到大排序，这个就可以保证，求和时，当前点的音量与前面比便是最大的。然后定义两个树状数组，一个记录这个点与前面点的距离，一个记录这个点前面点的个数。可以把这个位置放在一个数轴上，首先这个点前面的点的距离我们很好理解tree[i].a*(tree[i].bp-sum)，然后就是求后面的距离，你想1-5和5-1是一样的，因为1的音量在 5 6 后面，证明有两条路径 1-6，1-5，我们可以求总距离sum（20000，1），然后减去前面的距离和，然后减去后面的个数乘tree[i].b,便是与后面的距离差。tree[i].a(sum1(20000，1)-sum-(i-p-1)*tree[i].b)。比如算3这个位置时，3的坐标前面有1，后面有 5 6，我们知道的是 3-1，3-6，3-5。
  好好理解一下就好，上代码。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define lowbit(x) x&(-x)
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
ll t[maxn];
int s[maxn];
struct node
{
    int a,b;
}tree[maxn];
int cmp(node x,node y)
{
    if(x.a==y.a)
        return x.b<y.b;
    return x.a<y.a;
}
void init1(int x,int y)//记录的是与这个点的距离
{
    if(x==0)
        return;
    while(x<=maxn)
    {
        t[x]+=y;
        x+=lowbit(x);
    }
}
void init2(int x)//记录的是与这个点前面点的个数
{
    if(x==0)
        return;
    while(x<=maxn)
    {
        s[x]+=1;
        x+=lowbit(x);
    }
}
int sum1(int x)//距离的和
{
   ll ans=0;
    while(x>0)
    {
        ans+=t[x];
        x-=lowbit(x);
    }
    return ans;
}
int sum2(int x)//个数的和
{
    ll ans=0;
    while(x>0)
    {
        ans+=s[x];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
   int n;
   scanf("%d",&n);
   for(int i=1;i<=n;i++)
   {
       scanf("%d %d",&tree[i].a,&tree[i].b);
   }
   sort(tree+1,tree+n+1,cmp);
   ll ans=0;
    for(int i=1;i<=n;i++)
    {
        ll sum=sum1(tree[i].b);//距离
        int p=sum2(tree[i].b);//个数
        ans+=tree[i].a*(tree[i].b*p-sum)+tree[i].a*(sum1(20000)-sum-(i-p-1)*tree[i].b);//距离等于这个位置前面的数的距离和这个位置后面的距离，因为按照音量排序，所以这个点现在是最大的v
        init1(tree[i].b,tree[i].b);
        init2(tree[i].b);
    }
    printf("%lld\n",ans);
    return 0;
}