[树状数组] 树状数组基础练习题题解（三）

POJ-1990 MooFest

Time Limit: 1000MS
Memory Limit: 30000K

Contents

1 Description
2 Input
3 Output
4 Sample Input
5 Sample Output
6 Source

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

Sample Output

Source

USACO 2004 US Open

题目大意：每头牛有两个参数，一个坐标 $$x$$，一个音量的大小 $$v$$，求出两两间权值和 $$ max(v_i, v_j ) \times dis_{i->j} $$

解题思路：先按音量大小从小到大排序，编号为i的牛与编号小于 $$i$$ 的牛喊话，音量最大值即为 $$v_i$$ ,距离我们可以使用前缀和来计算。

维护两个树状数组，一个维护个数的前缀和，一个维护距离的前缀和。
令编号为 $$i$$ 的牛的左边（含）有 $$x$$ 头牛，右边有 $$y$$ 头牛（可用总数目 $$-x$$ ）。
与坐标在左边牛的权值和 = $$v_i * (x * dis_i – 前 x-1 头牛的权值和）$$
与坐标在右边牛的权值和 = $$v_i * (总权值和 – 前x头牛的权值和 – y * dis_i) $$

#include
#include
#include
#include
#include
#include
#define LL long long

using namespace std;

inline LL read(){
    LL x = 0;
    LL p = 1;
    char ch = getchar();

    while(ch < '0' || ch > '9'){
        if(ch == '-')
            p = -1;
        ch = getchar();
    }

    while('0' <= ch && ch <= '9'){
        x = x*10 + ch - '0';
        ch = getchar();
    }

    return x*p;
}

const int MAXN = 20000;

struct Data{
    LL v;
    LL x;
};

Data input[MAXN+10];

inline bool cmp(Data p,Data q){
    return p.v