MooFest
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 7077 | Accepted: 3181 |
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
【题意】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。
【分析】真想不到这题居然用到树状数组。首先将这n头牛按照v值从小到大排序（后面说的排在谁的前面，都是基于这个排序）。这样，排在后面的牛和排在前面的牛讲话，两两之间所用的音量必定为后面的牛的v值，这样一来才有优化的余地。然后，对于某头牛i来说，只要关心跟排在他前面的牛交流就好了。我们必须快速地求出排在他前面的牛和他之间距离的绝对值之和ans，只要快速地求出ans，就大功告成。这里需要两个树状数组。树状数组可以用来快速地求出某个区间内和，利用这个性质，我们可以快速地求出对于牛i，x位置比i小牛的个数，以及这个牛的位置之和。这里就需要两个树状数组，一个记录比x小的牛的个数sump，一个记录比x小的牛的位置之和sumx，然后，我们可以快速地求出牛i和比牛i位置小的牛的所有距离的绝对值为：sump*a[i].x-sumx;也可以方便地求出比牛i位置大的牛到牛i的距离和，即所有距离-sumx-(i-1-sump)*a[i].x；那么此题就差不多了。此题就用奶牛的坐标作为tree数组的下标，比较容易计算距离。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 2e4+10;
const int M = 24005;
const int mod=1e9+7;
ll tree[2][N];
int n,m;
struct man{
int x,v;
bool operator< (const man &it)const{
return v<it.v;
}
}a[N];
void add(int k,int num){
while(k<=20000){
tree[1][k]+=num;
tree[0][k]+=1;
k+=k&(-k);
}
}
ll Sum(int k,int d){
ll sum=0;
while(k>0){
sum+=tree[d][k];
k-=k&(-k);
}
return sum;
}
int main() {
scanf("%d",&n);
int v,x;
for(int i=1;i<=n;i++){
scanf("%d%d",&a[i].v,&a[i].x);
}
sort(a+1,a+1+n);
ll ans=0;
for(int i=1;i<=n;i++){
ll sump=Sum(a[i].x,0);
ll sumx=Sum(a[i].x,1);
ans+=a[i].v*(sump*a[i].x-sumx)+a[i].v*(Sum(20000,1)-sumx-(i-1-sump)*a[i].x);
add(a[i].x,a[i].x);
}
printf("%lld\n",ans);
return 0;
}