[贪心]Rest Stops

题目描述

Farmer John and his personal trainer Bessie are hiking up Mount Vancowver. For their purposes (and yours), the mountain can be represented as a long straight trail of length LL meters (1≤L≤10⁶). Farmer John will hike the trail at a constant travel rate of rF seconds per meter (1≤rF≤10⁶). Since he is working on his stamina, he will not take any rest stops along the way.
Bessie, however, is allowed to take rest stops, where she might find some tasty grass. Of course, she cannot stop just anywhere! There are N rest stops along the trail (1≤N≤10⁵); the i-th stop is xi meters from the start of the trail (0<xi<L) and has a tastiness value ci (1≤ci≤10⁶). If Bessie rests at stop i for t seconds, she receives ci⋅t tastiness units.

When not at a rest stop, Bessie will be hiking at a fixed travel rate of rB seconds per meter (1≤rB≤10⁶). Since Bessie is young and fit, rB is strictly less than rF.

Bessie would like to maximize her consumption of tasty grass. But she is worried about Farmer John; she thinks that if at any point along the hike she is behind Farmer John on the trail, he might lose all motivation to continue!

Help Bessie find the maximum total tastiness units she can obtain while making sure that Farmer John completes the hike.

输入

The first line of input contains four integers: L, N, rF, and rB. The next N lines describe the rest stops. For each i between 1 and N, the i+1-st line contains two integers xi and ci, describing the position of the i-th rest stop and the tastiness of the grass there.
It is guaranteed that rF>rB, and 0<x1<⋯<xN<L. Note that rF and rB are given in seconds per meter!

输出

A single integer: the maximum total tastiness units Bessie can obtain.

样例输入

10 2 4 3
7 2
8 1

样例输出

15

思路：总共可以休息的时间等于max_x*(rf-rb)；那么在每一个站休息时间的总和等于max_x*(rf-rb)；那么要使总val最大,则要对val值大的站分配更多一些的时间；

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;

struct STOP{
  ll pos,val;
  ll max_time;
}stop[100010];

bool cmp(STOP a,STOP b){
  return a.val>b.val;
}

int main()
{
    ll L,n,rf,rb;
    scanf("%lld%lld%lld%lld",&L,&n,&rf,&rb);
    for(ll i=1;i<=n;i++){
        scanf("%lld%lld",&stop[i].pos,&stop[i].val);
        stop[i].max_time=stop[i].pos*(rf-rb);
    }
    sort(stop+1,stop+1+n,cmp);
    ll past_time=0;
    ll ans=0;
    for(ll i=1;i<=n;i++){
        stop[i].max_time=(stop[i].max_time-past_time>0)?stop[i].max_time-tot_time:0;
        past_time+=stop[i].max_time;
        ans+=stop[i].max_time*stop[i].val;
    }
    printf("%lld
",ans);
    return 0;
}