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…

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…

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"…

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…

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…

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] +…

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.…

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…

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…

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…

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…

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…

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are…

迪杰斯特拉算法(Dijkstra):求一点到另外一点的最短距离 两种实现方法: 邻接矩阵,时间复杂度O(n^2) 邻接表+优先队列,时间复杂度O(mlogn)(适用于稀疏图) (n:图的节点数,m:图的边数) (参考 https://leetcode-cn.com/problems/path-with-maximum-probability/solution/gai-lu-zui-da-de-lu-jing-by-leetcode-solution/) leetcode经典例题: (1) 743…

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…

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…

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>…

本文主要是将我对于我对于迪杰斯特拉算法的理解写出来,同时通过例题来希望能够加深对于算法的理解,其中有错误的地方希望大家指正. 迪杰斯特拉算法 我将这个算法理解成一个局部到整体的算法,这个方法确实越研究就会发现越经典. 首先可以将整个图的节点看成两个集合:一个是S,一个是U-S.如果是求v0到图中各点的最短距离的话,那么S就是已经确认到v0距离最短的点,U-S则是对于整体的点集合U,还没有加入S集合的点. 这里提出一个算法总体的思想,将所有的点按照一定的原则加入到S集就是解集.而这个解法就是重点了…

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…

#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…

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…

                                                                                           1002 - Country Roads    PDF (English) Statistics Forum Time Limit: 3 second(s) Memory Limit: 32 MB 点击打开链接 I am going to my home. There are many cities and many…

分析:可以求简单的任意两点间最短距离的稍微变形,一个板子题.  #include <iostream> #include <bits/stdc++.h> using namespace std; int inf = 0x3fffff; int gra[1005][1005]; int mon[1005][1005]; int vis[1005]; int dist[1005]; int cost[1005]; void dijkstra(int s,int d,int n) { m…

单源最短路 所有边权都是正数 朴素Dijkstra算法(稠密图) #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int N=510; int n,m; int g[N][N]; int dist[N]; bool st[N]; int dijkstra(){ memset(dist,0x3f,sizeof…

1087 有多少不同的值(20 分) 当自然数 n 依次取 1.2.3.--.N 时,算式 ⌊n/2⌋+⌊n/3⌋+⌊n/5⌋ 有多少个不同的值?(注:⌊x⌋ 为取整函数,表示不超过 x 的最大自然数,即 x 的整数部分.) 输入格式: 输入给出一个正整数 N(2≤N≤10​4​​). 输出格式: 在一行中输出题面中算式取到的不同值的个数. 输入样例: 2017 输出样例: 1480 使用STL-set 的key进行索引 #include<iostream> #include<set&g…

1087 有多少不同的值(20 分) 当自然数 n 依次取 1.2.3.--.N 时,算式 ⌊n/2⌋+⌊n/3⌋+⌊n/5⌋ 有多少个不同的值?(注:⌊x⌋ 为取整函数,表示不超过 x 的最大自然数,即 x 的整数部分.) 输入格式: 输入给出一个正整数 N(2≤N≤10​4​​). 输出格式: 在一行中输出题面中算式取到的不同值的个数. 输入样例: 2017 输出样例: 1480 #include<iostream> #include<set> using namespace…

https://pintia.cn/problem-sets/994805260223102976/problems/1038429191091781632 当自然数 n 依次取 1.2.3.…….N 时,算式 ⌊ 有多少个不同的值?(注:⌊ 为取整函数,表示不超过 x 的最大自然数,即 x 的整数部分.) 输入格式: 输入给出一个正整数 N(2). 输出格式: 在一行中输出题面中算式取到的不同值的个数. 输入样例: 2017 输出样例: 1480 时间复杂度:$O(N)$ 代码: #inclu…

二级最短路+二级最短路,就是DP过程吧. 代码稍微注释一些,毕竟贴代码不好.. #include<bits/stdc++.h> using namespace std; typedef long long LL; const int INF=0x3f3f3f3f; const int N=4e4+10; struct asd{ int cost; int to; int next; }e[N]; int head[220],tol,tot; int n,m,w[220]; map<str…