poj 3431 Expedition 优先队列 题目链接: http://poj.org/problem?id=2431 思路: 优先队列.对于一段能够达到的距离,优先选择其中能够加油最多的站点,这样,行驶过这段距离之后还能走更远的距离. 将输入的数据进行排序处理,按照位置的先后.注意输入的距离是与终点的,要转化成与起点的. 代码: #include <iostream> #include <algorithm> #include <stdio.h> #include…

来源poj2431 A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every…

题目描述 A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit…

Description A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel ever…

题目链接 Description A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel…

Expedition 直接中文 Descriptions 一群奶牛抓起一辆卡车,冒险进入丛林深处的探险队.作为相当差的司机,不幸的是,奶牛设法跑过一块岩石并刺破卡车的油箱.卡车现在每运行一个单位的距离泄漏一个燃料单位. 为了修理卡车,奶牛需要沿着一条蜿蜒的长路行驶到最近的城镇(距离不超过1,000,000个单位).在这条路上,在城镇和卡车的当前位置之间,有N(1 <= N <= 10,000)燃料站,奶牛可以停下来获得额外的燃料(每站1..100个单位). 丛林对人类来说是一个危险的地方,对奶…

Expedition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10025   Accepted: 2918 Description A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed t…

题目链接: 传送门 Expedition Time Limit: 1000MS     Memory Limit: 65536K 题目描述 驾驶一辆卡车行驶L单位距离.最开始有P单位的汽油.卡车每开1单位距离消耗1单位的汽油.在途中一共有N个加油站.假设卡车的燃料箱容量无限大,问如果到达终点最少的加油次数. 思路 在卡车开往终点途中,只有在加油站才可以加油,换做认为"在到达加油站i时,就获得了一次在之后任何时候都可以加油的权利" #include<iostream> #in…

http://poj.org/problem?id=2431 你需要驾驶一辆卡车做一次长途旅行,但是卡车每走一单位就会消耗掉一单位的油,如果没有油就走不了,为了修复卡车,卡车需要被开到距离最近的城镇,在当前位置和城镇之间有n个加油站可以加油. 为了减少危险,需要最少的加油次数,卡车的油箱可以看作无限大,卡车初始距离城镇L单位,自身有P单位的油. 注意输入的距离是与城镇的距离不是与开始点的距离.转换一下就好. 思想:把经过的所有加油站都加入优先队列,当燃料不足时就取出优先队列的最大元素,用来给卡车…

思路:先把加油站按升序排列. 在经过加油站时.往优先队列里增加B[i].(每经过一个加油站时,预存储一下油量) 当油箱空时:1.假设队列为空(能够理解成预存储的油量),则无法到达下一个加油站,更无法到达目的地. 2.否则就取出队列里的最大元素,来给汽车加油(贪心思想) #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue>…