//贪心算法解决加油站选择问题 //# include<iostream> # include<stdio.h> using namespace std; # include<algorithm> struct Node { float p, d; }; bool cmp(Node a, Node b) { return a.d < b.d; } int main() { Node node[]; float Cmax, D, Davg, distance, pr…

题目1437:To Fill or Not to Fill 时间限制:1 秒 内存限制:128 兆 特殊判题:否 提交:1488 解决:345 题目描述: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way fro…

题目大意:小明从杭州去往某目的地,要经过一些加油站,每个加油站的价格不一样.若能顺利到达,求加油费用最少为多少,否则求出能行驶的最远距离. 思路:贪心算法 1>若下一加油站的价格更便宜,则只需走到下一加油站即可. 2>若下一结点的价格没有该节点便宜 1.若将油箱加满,看看在其能到达的最远距离内,是否有比该点更便宜的站点.若有,则正好到达这个跟便宜的点即可:否则,将油箱加满,然后到达这段距离内价格最小的点(除当前点外). 代码如下: #include<cstdio> #include…

 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 10 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 ZHANG, Guochuan With highways available, driving a car from Hangzhou to any other city is easy. But…

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, Guochuan With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to fin…

1033 To Fill or Not to Fill(25 分) With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may gi…

1033 To Fill or Not to Fill (25 分)   With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may…

Source: PAT A1033 To Fill or Not to Fill (25 分) Description: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. D…

1033 To Fill or Not to Fill With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give dif…

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked…

PAT A 1033 To Fill or Not to Fill With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may gi…

1033 To Fill or Not to Fill (25 分) With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may g…

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给不同的价格.您被要求仔细设计最便宜的路线. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行包含4个正数:Cmax(<= 100),坦克的最大容量; D(<= 30000),杭州与目的地城市的距离; Davg(<= 20),汽车可以运行的单位气体的平均距离;和N(<=…

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, Guochuan With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to fin…

题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1033 此题是一道贪心算法题,难度较大,关键在于贪心策略的选择: #include <cstdio> #include <vector> #include <algorithm> #include <climits> using namespace std; struct GasStation { double price; double distance…

转自:https://www.cnblogs.com/XBWer/p/3866486.html With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas…

题目如下: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are…

一.技术总结 是贪心算法的题目,题目主要考虑的问题有几个,是否会在第一个加油站的最近距离大于0,如果是这样那么直接输出答案,因为初始油箱没有汽油: 第二个是如何选定加油站,如果在可到达距离范围类,我们优先考虑比当前加油站价格更低的,然后如果有,就直接到达这里,如果没有那也要选出这里面价格最低的那个加油站, 然后在当前加油站,加满油箱.这样可以更加的省钱,那么油箱会多出油行驶距离(leftdis = Cmax*Davg -(minPriceDis - nowdis)). 可以开始判断最后结局了,如…

题目 With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are ask…

题目 本题的贪心算法策略需要深入思考一下 看到题目,最初没有理解题目的要求:看尽量多的完整的节目.尽量多是指数量多,自己理解成观看的时间最长.这样想其实简化了这道题. 正确理解题意后,首先想到的想法是:选择一个节目A,结束后选择另一个节目,如果节目A的时间同时覆盖了节目BC的时间,那么A就不应该看.怎么选择合适的节目?如果都把所有的节目考察一遍,统计看到的节目数量,成了穷举法,不是贪心算法. 遇到第一个关卡:如何在思维上找到突破口/切入点,当思路到达"如何选择合适的节目"这个问题时卡住…

题目描述: Yogurt factory The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <=…

题目描述: 给定一个非负整数数组,你最初位于数组的第一个位置. 数组中的每个元素代表你在该位置可以跳跃的最大长度. 判断你是否能够到达最后一个位置. 示例 1: 输入: [2,3,1,1,4] 输出: true 解释: 我们可以先跳 1 步,从位置 0 到达 位置 1, 然后再从位置 1 跳 3 步到达最后一个位置. 示例 2: 输入: [3,2,1,0,4] 输出: false 解释: 无论怎样,你总会到达索引为 3 的位置.但该位置的最大跳跃长度是 0,所以你永远不可能到达最后一个位置. 题…

题目描述: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are…

题目描写叙述: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You ar…

题目地址:http://ac.jobdu.com/problem.php?pid=1437 题目描述: With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different…

#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, id…

题目描写叙述 With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are…

题目 With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Diferent gas station may give diferent price. You are asked…

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked…

PAT-A最后一个问题.最后做出来... 贪婪,通过局部优化全局优化. 1. 该加油站按距离升序排列 2. 记录气体台当前所在index,目前的汽油.开支.在您的整个背部 3. 遍历中有两种情况: 1) 若发现油价比index更低的站next: 立即跳到该站(此时可能须要加油),不再继续遍历 -- 由于即使想要到达next后面的站,能够通过在next站购买更廉价的汽油来实现 2) 没有发现油价比index更低的站,则选择全部站中油价最低的站作为next: 此时考虑能否通过index抵达终点,若能…