[USACO08OPEN]牛的车Cow Cars

题目描述

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.

编号为1到N的N只奶牛正各自驾着车打算在牛德比亚的高速公路上飞驰．高速公路有M(1≤M≤N)条车道．奶牛i有一个自己的车速上限Si(l≤Si≤1,000,000)．

在经历过糟糕的驾驶事故之后，奶牛们变得十分小心，避免碰撞的发生．每条车道上，如果某一只奶牛i的前面有南只奶牛驾车行驶，那奶牛i的速度上限就会下降kD个单位，也就是说，她的速度不会超过Si – kD(O≤D≤5000)，当然如果这个数是负的，那她的速度将是0．牛德比亚的高速会路法规定，在高速公路上行驶的车辆时速不得低于/(1≤L≤1,000,000)．那么，请你计算有多少奶牛可以在高速公路上行驶呢？

输入输出格式

输入格式：

• 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

输出格式：

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

输入输出样例

输入样例#1： 复制

3 1 1 5 
5 
7 
5 

输出样例#1： 复制

2 

说明

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

从小到大排序，然后贪心选择车道；

代码

 1 #include
 2 #include
 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.dy.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 }

 

转载于:https://www.cnblogs.com/J-william/p/7801125.html