POJ1990 MooFest

Posted 2020-08-09 SilverNebula

 

Time Limit: 1000MS

Memory Limit: 30000KB

64bit IO Format: %lld & %llu

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

 

 

将牛按音量由小到大排序，算每头牛时，计算它和所有音量小于它的牛所贡献的答案。

用树状数组存牛的位置和坐标，由小到大依次添加。距离的计算方式见代码。

 

 

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #define LL long long
 8 using namespace std;
 9 const int mxn=30010;
10 struct node{
11     int v;
12     int x;
13 }c[mxn];
14 int cmpv(const node a,const node b){
15     return a.v<b.v;
16 }
17 int n;
18 LL ans=0;
19 //
20 LL t[2][mxn+1];
21 inline int lowbit(int x){return x&-x;}
22 void add(int k,int p,int v){//k==0 个数 k==1 距离 
23     while(p<=mxn){t[k][p]+=v;p+=lowbit(p);}
24 }
25 LL smm(int k,int x){
26     LL res=0;
27     while(x){res+=t[k][x];x-=lowbit(x);}
28     return res;
29 }
30 //
31 int main(){
32     scanf("%d",&n);
33     int i,j;
34     for(i=1;i<=n;i++) scanf("%lld%lld",&c[i].v,&c[i].x);
35     sort(c+1,c+n+1,cmpv);
36     for(i=1;i<=n;i++){
37         int num=smm(0,c[i].x);
38         LL dis=smm(1,c[i].x);
39         ans+=c[i].v*(num*c[i].x-dis);//算x小于当前值的牛 
40         ans+=(smm(1,mxn)-dis-(i-1-num)*c[i].x)*c[i].v;//算x大于当前值的牛 
41         add(0,c[i].x,1);
42         add(1,c[i].x,c[i].x);
43     }
44     printf("%lld\n",ans);
45     return 0;
46 }