PAT 1033. To Fill or Not to Fill
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <algorithm> using namespace std; int main() {
int N;
double mcap, dist, davg; scanf("%lf%lf%lf%d", &mcap, &dist, &davg, &N);
double price, idist; vector<pair<double, double> > stations(N + );
for (int i=; i<N; i++) {
scanf("%lf%lf", &price, &idist);
stations[i].first = idist;
stations[i].second= price;
} // virtual station to indicate the end
stations[N].first = dist;
stations[N].second= 1e300; sort(stations.begin(), stations.end()); double tank = , cur_dist = , cur_price = ; if (stations[].first > ) {
printf("The maximum travel distance = %.2lf", cur_dist);
return ;
} double money = ;
int cur_stat = ;
int cheap_stat = -;
double cheap_price = ; while (cur_dist < dist && (cur_stat) < N) {
double next_dist = stations[cur_stat + ].first;
double max_dist = mcap * davg + cur_dist;
// can't reach next station
if (max_dist < next_dist) {
cur_dist = max_dist;
break;
}
// can reach next station
// find first min gas price from cur_stat/cheap_stat within the reach range
if (cheap_stat < cur_stat) {
// last cheap station we calculated has past by
cheap_stat = cur_stat;
cheap_price= stations[cur_stat].second;
}
for (int i=cheap_stat; i<N && stations[i].first <= max_dist; i++) {
if (stations[i].second < cheap_price) {
cheap_stat = i;
cheap_price= stations[i].second;
break;
}
}
// cheaper station is cur_stat
if (cheap_stat == cur_stat) {
// from here we can reach end station
if (max_dist >= dist) {
money += cheap_price * (dist - cur_dist) / davg;
cur_dist = dist;
break;
}
// can't reach the end station
// we should fill full tank
money += (mcap - tank) * cheap_price;
tank = mcap;
} else {
// cur_stat is not the cheaper station,
// just fill enough gas, go to the cheaper one
double gas_need = (stations[cheap_stat].first - cur_dist) / davg;
if (gas_need > tank) {
// we have to fill gas here
money += (gas_need - tank) * stations[cur_stat].second;
tank = gas_need;
} else {
// we have the enough gas already
} }
// now we have enough gas
// drive to the next station
tank -= (next_dist - cur_dist) / davg;
cur_stat++;
cur_dist = next_dist;
}
if (cur_dist < dist) {
printf("The maximum travel distance = %.2lf", cur_dist);
} else {
printf("%.2lf", money);
}
return ;
}
写得有点烦
PAT 1033. To Fill or Not to Fill的更多相关文章
- PAT甲级1033. To Fill or Not to Fill
PAT甲级1033. To Fill or Not to Fill 题意: 有了高速公路,从杭州到任何其他城市开车很容易.但由于一辆汽车的坦克容量有限,我们不得不在不时地找到加油站.不同的加油站可能会 ...
- 【贪心】PAT 1033. To Fill or Not to Fill (25)
1033. To Fill or Not to Fill (25) 时间限制 10 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 ZHANG, Gu ...
- PAT 1033 To Fill or Not to Fill[dp]
1033 To Fill or Not to Fill(25 分) With highways available, driving a car from Hangzhou to any other ...
- PAT 甲级 1033 To Fill or Not to Fill (25 分)(贪心,误以为动态规划,忽视了油量问题)*
1033 To Fill or Not to Fill (25 分) With highways available, driving a car from Hangzhou to any oth ...
- PAT甲级——1033 To Fill or Not to Fill
1033 To Fill or Not to Fill With highways available, driving a car from Hangzhou to any other city i ...
- 1033 To Fill or Not to Fill
PAT A 1033 To Fill or Not to Fill With highways available, driving a car from Hangzhou to any other ...
- 1033. To Fill or Not to Fill (25)
题目链接:http://www.patest.cn/contests/pat-a-practise/1033 题目: 1033. To Fill or Not to Fill (25) 时间限制 1 ...
- 1033 To Fill or Not to Fill (25 分)
1033 To Fill or Not to Fill (25 分) With highways available, driving a car from Hangzhou to any other ...
- pat1033. To Fill or Not to Fill (25)
1033. To Fill or Not to Fill (25) 时间限制 10 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 ZHANG, Gu ...
- PAT_A1033#To Fill or Not to Fill
Source: PAT A1033 To Fill or Not to Fill (25 分) Description: With highways available, driving a car ...
随机推荐
- [转] 配置文件现在需要绝密的短语密码(blowfish_secret)的解决方法
今天在使用 phpMyAdmin 操作数据库时,刚刚登陆后发现最下面有如下信息提示: 配置文件现在需要绝密的短语密码(blowfish_secret). 园子在网上找了多种解决方法,写的都不是非常详细 ...
- java面向对象之封装性
面向对象的特征之一:封装和隐藏: 一.问题的引入: 当创建一个类的对象以后,可以通过“对象.属性”的方式,对对象的属性进行赋值.这里,赋值操作要受属性的数据类型和存储范围的制约.除此之外,没有其他制约 ...
- 固定定位 z-index iconfont的使用 043
固定定位 现象 脱标 提升层级 将盒子固定住 参考点 浏览器的左上角 : <!DOCTYPE html> <html lang="en"> <he ...
- Partition Array by Odd and Even
Partition an integers array into odd number first and even number second. Example Given [, , , ], , ...
- .Net C# 泛型序列化和反序列化JavaScriptSerializer
项目添加引用System.Web.Extensions /// <summary> /// 泛型序列化 /// </summary> public class JsonHelp ...
- 自动化测试接口PYTHON
在开发测试中经常会遇到接口迭代和代码重构,一个无关紧要的改动往往会引起整个项目的运行.现有的接口测试中往往只是针对单一接口的测试,可是业务的连贯性是非常紧密的,比如:用户从登陆,获取商品信息,下单 ...
- erlang随笔3--OTP
OTP最核心的概念就是行为.一个行为封装了某种常见的行为模式.可以把这些行为理解为某种应用程序框架.可以通过回调模块来 定制这些框架.OTP依靠行为引用了容错,扩容和动态代码升级等特性.所以在写回调模 ...
- JS之作用域与闭包
JS之作用域与闭包 作用域在JS中同样也是一个重要的概念.它不复杂,因为ES5中只有全局作用域和函数作用域,我们都知道他没有块级作用域.但在ES6中多了一个let,他可以保证外层块不受内层块的影响 ...
- myEclipse或eclipse复制或修改项目后的部署名称
第一种方式: 右键工程-properties-Deployment-Deployment Assembly下 修改Web Content Root名称即可.见下图 第二种方式: 右键工程-proper ...
- pip升级最新版本
1.如果是python2.7输入以下指令 python -m pip install --upgrade pip 2.如果是python 3.+输入以下指令 python3 -m pip instal ...