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MooFest
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 7716 Accepted: 3474
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.
OutputLine 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
题意：
有n只牛在数轴上排成一排，每只牛都在两个属性v,x
v表示说话的声音，x表示牛的位置
当两只牛i,j要进行交谈时,他们产生的声音是 dis(xi,xj)*max(vi,vj) : dis表示两个值的差的绝对值
问你要是任意两只牛进行交谈，他们总共产生的声音为多少？
分析：
由于两只牛说话的声音是max(vi,vj)，所以我们可以按照v的升序来将牛排序，那么就可以保证当牛i和他前面的牛进行交谈时，是以vi来交谈的
我们现在只需要求出i和在他前面的所有的牛的位置的差值的总和就可以
了
我们将前面的牛分为两类：在前面且x比xi小的，和在前面x不小于xi的
我们假设
cnt为前面x比xi小的牛的个数
presume为前面x比xi小的牛的x的总和
total[i-1]表示前面i-1只牛的x的总和
那么i牛和在它前面且x比xi小的牛的差值的总和为 cnt * xi - presum
并且我们知道前面x不比xi小的牛的个数为 i-1 - cnt
i牛和在它前面且x不比xi小的牛的差值的总和为 total[i-1] - presum - (i-1 - cnt) * xi)
所以牛i产生的声音为 cnt * xi - presum + total[i-1] - presum - (i-1 - cnt) * xi) * vi
所以我们只需要用树状数组来维护 cnt 和 presume 即可
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;const int maxn = 20010;
int c[maxn],d[maxn],total[maxn];
int n;struct Node
{int v,x;
}node[maxn];int lowbit(int x)
{return x & (-x);
}void update(int x)
{int t = x;while( x <= maxn){c[x]++;d[x] += t;x += lowbit(x);}
}int getsum_c(int x)
{int sum = 0;while(x > 0){sum += c[x];x -= lowbit(x);}return sum;
}int getsum_d(int x)
{int sum = 0;while(x >0){sum += d[x];x -= lowbit(x);}return sum;
}int cmp(Node a,Node b)
{return a.v < b.v;
}int main()
{scanf("%d",&n);memset(c,0,sizeof(c));memset(d,0,sizeof(d));memset(total,0,sizeof(total));for(int i=0;i<n;i++) scanf("%d%d",&node[i].v,&node[i].x);sort(node,node+n,cmp);long long ans = 0;for(int i=0;i<n;i++){if(i == 0){total[i] = node[i].x;update(node[i].x);}else{total[i] = total[i-1] + node[i].x;int cnt = getsum_c(node[i].x);int presum = getsum_d(node[i].x);ans += (1LL * node[i].v * (cnt * node[i].x - presum + total[i-1] - presum - (i - cnt) * node[i].x));update(node[i].x);}}printf("%lld",ans);return 0;
}
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