1072. Gas Station (30) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. H…

1072 Gas Station (30 分)   A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service r…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of…

题目 A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map…

1072 Gas Station (30)(30 分) A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service…

题目信息 1072. Gas Station (30) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of…

先要求出各个加油站 最短的 与任意一房屋之间的 距离D,再在这些加油站中选出最长的D的加油站 ,该加油站 为 最优选项 (坑爹啊!).如果相同D相同 则 选离各个房屋平均距离小的,如果还是 相同,则 选 编号小的. 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A gas station has to be built at such a location that the minimum distance…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. 考试的时候这句话意思没有完全理解,悲剧鸟... 其实就是让你找一个加油站,它到所有房子最短距离为len , 找一个加油站len的距离越大越好 加油站最多10个,以这几个加油站为原点,做…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of…

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range. Now given the map of…

题意: 输入四个正整数N,M,K,D(N<=1000,M<=10,K<=10000)分别表示房屋个数,加油站个数,路径条数和加油站最远服务距离,接着输入K行每行包括一条路的两条边和距离.输出最近距离最远同时与所有房屋距离都不超过D的加油站名字和最近距离以及平均距离.如果有多个答案则优先输出平均距离最小的,再不唯一则输出编号最小的. trick: 这题有一个诡异的事情,样例1的平均距离是3.25,取一位小数应该是3.2,四舍五入则是3.3,样例显示为3.3,此时我理解为四舍五入,题面说是精…

枚举一下选的位置,每次算一下就可以了. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; const int…

题意:从m个加油站里面选取1个站点,使得其离住宅的最近距离mindis尽可能地远,并且离所有住宅的距离都在服务范围ds之内.如果有很多相同mindis的加油站,输出距所有住宅平均距离最小的那个.如果平均值还是一样,就输出按照顺序排列加油站编号最小的. 分析:加油站之间也是彼此有路连接的,所以最短路径计算的时候也要把加油站算上,是双向边,所以边的数组大小得开两倍.加油站编号记为n+1~n+m,对这些点用dijkstra计算最短路径即可,这里顺便在dijkstra中进行了剪枝处理,不剪枝也没事. #…

#include<bits/stdc++.h> using namespace std; ; int n,m,k,Ds; int mp[N][N]; int dis[N]; int vis[N]; int inf=0x3f3f3f3f; int toint(char s[]) { ; int len=strlen(s); ;i<len;i++){ if(s[i]=='G') continue; sum=sum*+s[i]-'; } ]=='G') return sum+n; return…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…

本文为PAT甲级分类汇编系列文章. 图,就是层序遍历和Dijkstra这一套,#include<queue> 是必须的. 题号 标题 分数 大意 时间 1072 Gas Station 30 最短距离最大,距离不超限 200ms 1076 Forwards on Weibo 30 一定层数的转发计数 3000ms 1079 Total Sales of Supply Chain 25 计算供应链总销售额 250ms 1087 All Roads Lead to Rome 30 复杂权重的最短路…

今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是modern C++. 为什么要整理呢,因为我2019年9月要考PAT甲级,虽然是第一次考,虽然只学了数据结构(https://mooc.study.163.com/course/1000033001?tid=2402970002#/info),但我要冲着高分(2019年9月8日更新:满分)去. 下…

2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

1072. Gas Station (30) 时间限制  200 ms 内存限制  32000 kB 代码长度限制  16000 B 判题程序    Standard 作者    CHEN, Yue A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as po…

   转载请注明出处:https://www.cnblogs.com/jlyg/p/7525244.html 终于在考前,刷完PAT甲级131道题目,不容易!!!每天沉迷在刷题之中而不能超脱,也是一种境界.PAT甲级题目总的说卡题目的比较多,卡测试点的比较少,有些题目还会有题意混淆,这点就不吐槽了吧.静下心来耍这130道题,其实磨练的是一种态度与手感,养成的是一种习惯.热爱AC没有错!!(根据自己的刷题情况持续更新中,这次是第二遍刷了,发现换了新网站后,oj严格很多了,有些之前可以过的,现在已经…

PAT甲级1131. Subway Map 题意: 在大城市,地铁系统对访客总是看起来很复杂.给你一些感觉,下图显示了北京地铁的地图.现在你应该帮助人们掌握你的电脑技能!鉴于您的用户的起始位置,您的任务是找到他/她目的地的最快方式. 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行包含正整数N(<= 100),地铁线数.然后N行跟随,第i(i = 1,...,N)行以格式描述第i条地铁线: M S [1] S [2] ... S [M] 其中M(<= 100)是停止次数,S [i…

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(<=…

PAT甲级1018. Public Bike Management 题意: 杭州市有公共自行车服务,为世界各地的游客提供了极大的便利.人们可以在任何一个车站租一辆自行车,并将其送回城市的任何其他车站. 公共自行车管理中心(PBMC)不断监测所有车站的实时能力. 如果一个车站正好半满,那么一个车站据说处于完美的状态.如果车站充满或空,PBMC将收集或发送自行车,以调整该车站的状况.此外,所有的车站也将进行调整. 当报告问题站时, PBMC将始终选择到达该站的最短路径.如果有多于一条最短路径,则需要…

Source: PAT A1072 Gas Station (30 分) Description: A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the hou…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an e…