[USACO08OPEN]牛的車Cow Cars

題目描述

N (1 <= N <= 50,000) cows conveniently numbered 1..N are driving in separate cars along a highway in Cowtopia. Cow i can drive in any of M different high lanes (1 <= M <= N) and can travel at a maximum speed of S_i (1 <= S_i <= 1,000,000) km/hour.

After their other bad driving experience, the cows hate collisions and take extraordinary measures to avoid them. On this highway, cow i reduces its speed by D (0 <= D <= 5,000) km/hour for each cow in front of it on the highway (though never below 0 km/hour). Thus, if there are K cows in front of cow i, the cow will travel at a speed of max[S_i - D * K, 0]. While a cow might actually travel faster than a cow directly in front of it, the cows are spaced far enough apart so crashes will not occur once cows slow down as

described,

Cowtopia has a minimum speed law which requires everyone on the highway to travel at a a minimum speed of L (1 <= L <= 1,000,000) km/hour so sometimes some of the cows will be unable to take the highway if they follow the rules above. Write a program that will find the maximum number of cows that can drive on the highway while obeying the minimum speed limit law.

編號爲1到N的N只奶牛正各自駕着車打算在牛德比亞的高速公路上飛馳．高速公路有M(1≤M≤N)條車道．奶牛i有一個本身的車速上限Si(l≤Si≤1,000,000)．

在經歷過糟糕的駕駛事故以後，奶牛們變得十分當心，避免碰撞的發生．每條車道上，若是某一隻奶牛i的前面有南只奶牛駕車行駛，那奶牛i的速度上限就會降低kD個單位，也就是說，她的速度不會超過Si – kD(O≤D≤5000)，固然若是這個數是負的，那她的速度將是0．牛德比亞的高速會路法規定，在高速公路上行駛的車輛時速不得低於/(1≤L≤1,000,000)．那麼，請你計算有多少奶牛能夠在高速公路上行駛呢？

輸入輸出格式

輸入格式：

• Line 1: Four space-separated integers: N, M, D, and L

• Lines 2..N+1: Line i+1 describes cow i's initial speed with a single integer: S_i

輸出格式：

• Line 1: A single integer representing the maximum number of cows that can use the highway

輸入輸出樣例

輸入樣例#1： 複製

3 1 1 5 
5 
7 
5 

輸出樣例#1： 複製

2 

說明

There are three cows with one lane to drive on, a speed decrease of 1, and a minimum speed limit of 5.

Two cows are possible, by putting either cow with speed 5 first and the cow with speed 7 second.

思路

從小到大排序，而後貪心選擇車道；

代碼

 1 #include<cstdio>
 2 #include<algorithm>
 3 #define LL long long
 4 const int maxn=1e5+10;
 5 LL n,m,ans;
 6 struct nate{LL t,d;}s[maxn];
 7 bool comp(nate x,nate y){return x.t*y.d<x.d*y.t;}
 8 int main(){
 9     scanf("%lld",&n);
10     for(int i=1;i<=n;i++){
11         scanf("%lld%lld",&s[i].t,&s[i].d);
12         s[i].t*=2,m+=s[i].d;
13     }
14     std::sort(s+1,s+n+1,comp);
15     for(int i=1;i<=n;i++){
16         m-=s[i].d;
17         ans+=s[i].t*m;
18     }
19     printf("%lld\n",ans);
20     return 0;
21 }