poj 1990 MooFest（樹狀數組之代碼）

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

 

題目大意：一群牛參加完牛的節日後都有了不同程度的耳聾，第i頭牛聽見別人的講話，別人的音量必須大於v[i]，當兩頭牛i，j交流的時候，交流的最小聲音為max{v[i],v[j]}*他們之間的距離。現在有n頭牛，求他們之間兩兩交流最少要的音量和。

解題思路：首先要根據音量f進行排序，這樣就可以進行優化，即對於某一頭牛i來說，排在他前面的音量都比他小，肯定是用i自身的音量*兩者之間的距離。此時的計算公式為：（求解比當前x小的所有和）

(x-x1)*f;

(x-x2)*f;

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

綜上為：(n*x（x1+x2+x3+.....）)*f； 注釋：x表示某一頭牛當前的位置，x1，x2，x3......等表示排在他前面的牛的位置，f當前這頭牛的音量，因為他是最大的音量。（這裡的距離我們需要采用樹狀數組進行求解）

這裡有個地方需要解釋一下，就是排在這個牛i前面的音量一定都比他小，但是坐標值x不一定比他小，所以我們只能分開處理。

我們可以在建立一個數軸，一個點一個點的放進去，先放進去的音量肯定是小的，把他按照該有的固定位置放進去，這樣排在他前面的所有x都比他小就可以按照上面的方法采用樹狀數組的方法進行區間求和。（這裡要放入一個求解一次他左邊和右邊的值，左邊就是比當前x小的，右邊是比當前x大的）。

下面介紹比當前值x大的情況計算方法：

 

(x1-x)*f;

(x2-x)*f;

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

綜上：((x1+x2+x3+......)-n*x)*f;

這裡的n值就應該等於放進去的所有牛的個數-前面已經計算過的比他x小的個數：i-nn；

（x1+x2+x3+...）就應該等於總和減去比他小的x的和。

要用long long ！！！

這樣講起來很復雜，詳情請看代碼實現~~~~不懂得評論

 

詳見代碼。

 

#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