Poj1990 MooFest

最新推荐文章于 2021-04-14 16:10:12 发布

verdin黄大锤

最新推荐文章于 2021-04-14 16:10:12 发布

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本文链接：https://blog.youkuaiyun.com/rabbit_ZAR/article/details/69676852

本文探讨了一道经典的算法问题，即如何计算一群沿直线排列的牛为了相互沟通而产生的总体声音强度。通过将问题转化为特定的数据结构操作，如树状数组的应用，有效地解决了这一难题。

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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 U S Open

题目大意：有n头牛，排列成一条直线，给出每头牛的坐标x。每头牛都会有一个声音的大小值v，两头距离为d的牛a和b想说话，就需要用d*max(a.v,b.v)的声音。输出当所有牛都可以互相沟通时的声音大小。

思路：

先把牛按站的位置排序，只考虑每头牛和它前面的牛说话。

排好序后：

这时，求第i头牛和它前面的牛说话时发出的声音大小时，距离就可以用树状数组求了。

但是求声音大小时，要分向左向右两种情况考分别考虑：

这样，向两边的v也可以分别用一个树状数组求了。

向左：

for(int i=1;i<=n;i++){
	long long x= sum(cow[i].v-1, Num);
        long long y= sum(cow[i].v-1, Sum);
        S+= ((x*cow[i].x-y)*cow[i].v);
        add(cow[i].v, Num, 1);
        add(cow[i].v, Sum, cow[i].x);
}

向右：

for(int i=n;i>=1;i--){
	long long x= sum(cow[i].v, Num);
        long long y= sum(cow[i].v, Sum);
        S+= ((y-x*cow[i].x)*cow[i].v);
        add(cow[i].v, Num, 1);
        add(cow[i].v, Sum, cow[i].x);
}

注意：要用long long 型！