Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contains one test case. For each case, th…

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contains one test case. For each case, th…

Source: PAT A1087 All Roads Lead to Rome (30 分) Description: Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specificati…

1087 All Roads Lead to Rome (30)(30 分) Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file co…

1087. All Roads Lead to Rome (30) Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contain…

PAT 1087 All Roads Lead to Rome 题目: Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file conta…

PAT甲级1087. All Roads Lead to Rome 题意: 确实有从我们这个城市到罗马的不同的旅游线路.您应该以最低的成本找到您的客户的路线,同时获得最大的幸福. 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2个正整数N(2 <= N <= 200),城市数,K, 双城之间的路线总数;其次是起始城市的名称.下一个N-1行每个都给出一个城市的名字和一个整数,代表从城市可以获得的幸福,除了起始城市.然后K行跟随,每个描述两个城市之间的路线,格式为"C…

PAT1087. All Roads Lead to Rome 题目大意 给定一个图的边权和点权, 求边权最小的路径; 若边权相同, 求点权最大; 若点权相同, 则求平均点权最大. 思路 先通过 Dijkstra 求得最短路径, 需要注意的是: 要保证每次松弛时 u 和 v 不相同, 否则会形成自环, 则从 ROM 开始 BFS 遍历每一条边权相同的路径. 代码 #include <iostream> #include <cstdio> #include <vector>…

1087. All Roads Lead to Rome (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 Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gain…

题目链接 All Roads Lead to Rome 题目大意:求符合题意(三关键字)的最短路.并且算出路程最短的路径有几条. 思路:求最短路并不难,SPFA即可,关键是求总路程最短的路径条数. 我们令$h[i][j]$为$i$到$j$的最短路,$g[i][j]$为$i$到$j$的最短路条数. $f[i][j]$的求法就是常规的Floyd求法 $g[i][j]$利用乘法原理的性质来求. 当枚举的中间点$k$满足 $f[i][k] + f[k][j] = f[i][j]$时 $g[i][j] +…

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 Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Spec…

https://pintia.cn/problem-sets/994805342720868352/problems/994805379664297984 Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.…

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contains one test case. For each case, th…

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file contains one test case. For each case, th…

思路:单源最短路末班就好了,字符串映射成数字处理. AC代码 //#define LOCAL #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <string> using namespace std; #define inf 0x3f3…

直接用Dijkstra做 #include<bits/stdc++.h> using namespace std; int n,m; map<string,int>si; map<int,string>is; ; int weight[N]; int mp[N][N]; const int inf=0x3f3f3f3f; int dis[N]; bool vis[N]; int num[N]; int w[N]; int past[N]; int path[N]; vo…

#include<bits/stdc++.h>using namespace std;map<string,int>city;map<int,string>rcity;map<int,vector<pair<int,int> > >edge;//对比string要比对比int慢很多,所以转换映射int dis[207],path[207],hcount[207],happ[207],fstep[207],f[207];//源点到各点的…

暴力DFS. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<iostream> #include<algorithm> using namespace st…

题意: 输入两个正整数N和K(2<=N<=200),代表城市的数量和道路的数量.接着输入起点城市的名称(所有城市的名字均用三个大写字母表示),接着输入N-1行每行包括一个城市的名字和到达该城市所能获得的快乐,接着输入M行每行包括一条道路的两端城市名称和道路的长度.输出从起点城市到目标城市"ROM"获得最大快乐且经历道路长度最短的道路条数和经历的道路长度和获得的快乐总数以及其中经过城市最少的那条路的到达每个城市所获得的平均快乐.下一行输出这条路的路径,城市名之间用"…

题目分析: 这题我在写的时候在PTA提交能过但是在牛客网就WA了一个点,先写一下思路留个坑 这题的简单来说就是需要找一条最短路->最开心->点最少(平均幸福指数自然就高了),由于本题给出的所有的城市的名称都是字符串(三个大写字母),所以我先将每个字符串计算一个hash值去代表每一个城市,接着就是用dijkstra算法去逐步判断各种情况,是直接用了一个dijkstra去计算,而没有说是先dijkstra算出最短距离再dfs所有的路径去找出最优解,因为我觉得这题局部最优就能反映到全局最优,也不知是…

Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性表:栈,队列,链表,顺序表 树:二叉树的建立,二叉树的遍历,完全二叉树,二叉查找树,平衡二叉树,堆,哈夫曼树 图:图的存储和遍历 经典高级算法: 深度优先搜索,广度优点搜索,回溯剪枝 贪心,并查集,哈希映射 最短路径(只考察过单源),拓扑排序(18年9月第一次涉及相关概念,未正式考过),关键路径(未…

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"…

问题 A: 木棒根数 解法:把所有的情况保存下来,加一下就好 #include<bits/stdc++.h> using namespace std; map<char,int>q; class P { public: int cmd(string s) { ; ;i<s.length();i++) { sum+=q[s[i]]; } return sum; } }; int main() { string s; P solve; q[; q[; q[; q[; q[; q[…

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…

很多博友看了我的第一篇博客yolo类检测算法解析——yolo v3,对其有了一定的认识和了解,但是并没有贴出代码和运行效果,略显苍白.因此在把篇博客理论的基础上,造就了第一篇实践文章,也就是本文.只要读者有着强大的理论支撑,什么模型什么框架都是一样玩.所以老师又会跟你说哲学和科学.科学和技术存在如何的关联,尽管很抽象,但是没有人反驳过就像有这么多的编程语言,各有特色,最后也都干了相同或相似的事,那么多的框架,各有千秋,最后也都干了相同或相似的事.又或者说反了,是因为它们都想干相同或相似的事,而又…

Around 2009 the Stratosphere research project started at the TU Berlin which a few years later was set to become the Apache Flink project. Often compared with Apache Spark in addition to that Apache Flink offers pipelining (inter-operator parallism)…

9月台风席卷全球,本刊特别选取台风最佳可视化案例,数据可视化应用功力最深厚者,当属纽约时报,而传播效果最佳的是The Weather Channel关于Florence的视频预报,运用了数据可视化.可视化特效以及虚拟特效制作等技术,简直是灾难大片电影预告(在本文末尾查看视频). See Flood Waters Rise Across the Carolinas After Hurricane Florence Hurricane Florence brought a lot of rain,…

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…

PAT甲级题目:点这里 pat解题列表 题号 标题 题目类型  10001 1001 A+B Format (20 分)  字符串处理  1003 1003 Emergency (25 分) 最短路径(Dijkstra or spfa)  1013 1013 Battle Over Cities (25 分) 图的遍历or并查集  1018 1018 Public Bike Management (30 分) 最短经 and 图的遍历       1030 1030 Travel Plan (3…

近期一次pat考试中的最后一题.事实上玩算法这东西就像打魔兽.不能光有思想上的高度,微操必须实打实.就这么个迪杰斯特拉算法.多少教科书上都讲烂了. 可是现场又有多少人是敲对的呢?不能光停留在理解上.必须能用自己的方式表达出来才算过关. 题目: 1087. All Roads Lead to Rome (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 Indeed there are many diff…