poj 1990 MooFest（树状数组）

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6265		Accepted: 2765

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

题目大意：一群牛参加完牛的节日后都有了不同程度的耳聋，第i头牛听见别人的讲话，别人的音量必须大于v[i]，当两头牛i，j交流的时候，交流的最小声音为max{v[i],v[j]}*他们之间的距离。现在有n头牛，求他们之间两两交流最少要的音量和。

解题思路：首先要根据音量f进行排序，这样就可以进行优化，即对于某一头牛i来说，排在他前面的音量都比他小，肯定是用i自身的音量*两者之间的距离。此时的计算公式为：（求解比当前x小的所有和）

(x-x1)*f;

(x-x2)*f;

(x-x3)*f........

综上为：(n*x（x1+x2+x3+.....）)*f； 注释：x表示某一头牛当前的位置，x1，x2，x3......等表示排在他前面的牛的位置，f当前这头牛的音量，因为他是最大的音量。（这里的距离我们需要采用树状数组进行求解）

这里有个地方需要解释一下，就是排在这个牛i前面的音量一定都比他小，但是坐标值x不一定比他小，所以我们只能分开处理。

我们可以在建立一个数轴，一个点一个点的放进去，先放进去的音量肯定是小的，把他按照该有的固定位置放进去，这样排在他前面的所有x都比他小就可以按照上面的方法采用树状数组的方法进行区间求和。（这里要放入一个求解一次他左边和右边的值，左边就是比当前x小的，右边是比当前x大的）。

下面介绍比当前值x大的情况计算方法：

(x1-x)*f;

(x2-x)*f;

(x3-x)*f........

综上：((x1+x2+x3+......)-n*x)*f;

这里的n值就应该等于放进去的所有牛的个数-前面已经计算过的比他x小的个数：i-nn；

（x1+x2+x3+...）就应该等于总和减去比他小的x的和。

要用long long ！！！

这样讲起来很复杂，详情请看代码实现~~~~不懂得评论

详见代码。

#include 
#include 
#include 
#include 

using namespace std;

#define ll long long

struct node
{
    ll f,x;
} s[20010];

int n;
ll c1[20010],c2[20010];
ll a1[20010],a2[20010];

bool cmp(node a,node b)
{
    return a.f