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…
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"…
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…
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…
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…
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) 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…
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…
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.…
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…
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…
我先在CSDN上面发表了同样的文章,见https://blog.csdn.net/weixin_44385565/article/details/88863693 排版比博客园要好一些.. 1135 Is It A Red-Black Tree (30 分) There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 proper…
题意: 输入一个正整数N(<=1000),接着输入N个整数([-1000,1000]),依次插入一棵初始为空的二叉排序树.输出最底层和最底层上一层的结点个数之和,例如x+y=x+y. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct Node{ int value; int vis; int level; int visl,…
题意: 输入一个正整数N(<=100),接着输入N行每行包括0~N-1结点的左右子结点,接着输入一行N个数表示数的结点值.输出这颗二叉排序树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; pair<]; ]; ; ]; void dfs(int x){ ) return ; dfs(a[x].first); ans[x]=b[++c…
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>…
#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…
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…
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 A+B for Polynomials (25)(25 分) This time, you are supposed to find A+B where A and B are two polynomials. Input Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a~N1…
1043 Is It a Binary Search Tree (25 分)   A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a…
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…
There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 properties: (1) Every node is either red or black. (2) The root is black. (3) Every leaf (NULL) is black. (4) If a node is red, then bot…
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with…
题意: 输入一个正整数N(<=10000),接着依次输入N个学生的ID.输入一个正整数Q,接着询问Q次,每次输入一个学生的ID,如果这个学生的ID不出现在之前的排行榜上输出Are you kidding,否则如果已经询问过输出Checked,否则如果这位学生排名第一输出Mystery Award,否则如果这位学生的排名是质数输出Minion,否则输出Chocolate. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/…
题意:输入一个正整数N(<=1e5),两个小数P和R,分别表示树的结点个数和商品原价以及每下探一层会涨幅的百分比.输出叶子结点深度最小的商品价格和深度最小的叶子结点个数. trick: 测试点1只有根节点一个点,输出P和1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; vector<]; ]; void dfs(int x,int y){ )…