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<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 }