POJ_1990_MooFest_树状数组

MooFest

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7754 Accepted: 3500

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

有 N 头牛成一列，每头牛给定坐标 x 和 v。两头牛之间建立联系，代价是较大的 v * 它俩之间的距离。求任意一头牛都与其他所有牛建立联系的代价之和。

一开始想的是按照 v 从大到小的顺序遍历牛，这样这头牛的花费为 v * 它到剩下的其他牛的距离之和。然后就是怎样快速地求出它到其它牛的距离之和。但是没想出啥好办法。

后来想到按照 v 从小到大的顺序遍历牛， 这样这头牛的花费就是 v * 它到之前的牛的距离之和。求和的方法很关键，前一种思路就是因为没想到这个方法没做出来。算距离的方法是算距离之差。把 x 提出来，就可以算区间距离了。但是这样就不能加绝对值，要考虑正负问题。这就要把已经有的牛分成两部分，x 大的和 x 小的。在提 x 出来的过程中，我们需要知道有多少个比 x 小的（进而也知道了有多少个比 x 大的）。求区间和的问题考虑用树状数组解决。

维护两个树状数组。一个维护的是对应位置上牛的坐标值，用来求已有牛的坐标和；一个维护的是对应位置上是否有牛，用来求比 x 小的牛的数量。

#include<cstdio>#include<iostream>#include<algorithm>#define f(x) p[x].first#define s(x) p[x].secondusing namespace std;typedef long long LL;typedef pair<LL, LL> P;int N;const int maxn = 20000 + 10;const int inf  = 20000;//pos最大值P p[maxn];LL A[maxn];//times小且pos小的牛的pos和LL B[maxn];//times小且pos小的牛的数量void Add(LL *Bit, int i, int x){while(i <= inf){Bit[i] += x;i += i&-i;}}LL Sum(LL *Bit, int i){LL sum = 0;while(i){sum += Bit[i];i -= i&-i;}return sum;}bool cmp(P x, P y){return x.first < y.first;}int main(){scanf("%d", &N);//f是times，s是posfor(int i= 1; i<= N; i++)scanf("%lld %lld", &f(i), &s(i));//times由小到达排序sort(p+1, p+1+N, cmp);LL res = 0;for(int i= 1; i<= N; i++){//当前pos小于s(i)的pos和/数量LL a = Sum(A, s(i)), b = Sum(B, s(i));//前面牛res += (s(i) * b - a) * f(i);//后面牛res += (Sum(A, inf) - a - s(i) * (i-1-b)) * f(i);//把这个牛加入树状数组Add(A, s(i), s(i));Add(B, s(i), 1);}cout << res << endl;return 0;}