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POJ1990 MooFest

2024-08-13 22:22:53   215人阅读

 
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

43 12 52 64 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 }

 

POJ1990 MooFest


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