(一)题意 题目链接:https://www.patest.cn/contests/pat-a-practise/1030 1030. Travel Plan (30) A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a tra…

PAT 1030 最短路最小边权 堆优化dijkstra+DFS 1030 Travel Plan (30 分) A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest…

#include <iostream> #include <limits> #include <vector> using namespace std; int n,m,s,d; int cityMap[500][500]; int costMap[500][500]; #define INF numeric_limits<int>::max() int dp[500]; int vis[500]; int costDp[500]; vector<in…

1030. Travel Plan (30) A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting cit…

1030 Travel Plan (30 分)   A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting…

//晴神模板,dij+dfs,貌似最近几年PAT的的图论大体都这么干的,现在还在套用摸板阶段....估计把这及格图论题题搞完,dij,dfs,并查集就掌握差不多了(模板还差不多)为何bfs能自己干出来,dfs就各种跪....感觉需要把图论的经典算法都码一遍,才能有更深的理解,现在只是浅表 #include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace st…

时间限制 400 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 traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide t…

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination.…

题目 A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destinatio…

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination.…

https://www.patest.cn/contests/pat-a-practise/1030 找最短路,如果有多条找最小消耗的,相当于找两次最短路,可以直接dfs,数据小不会超时. #include<cstdio> #include<string> #include<cstring> #include<vector> #include<iostream> #include<queue> #include<bitset&g…

先处理出最短路上的边.变成一个DAG,然后在DAG上进行DFS. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<queue> #include<vector> using namespace std; const int INF=0x7FFFFFFF; ; int n,m,…

模板题最短路+输出路径如果最短路不唯一,输出cost最小的 #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> #include <queue> #define INF 0x3f3f3f3f using namespace std; ; int n,m,s,t; struct Edge{ int…

题意: 输入N,M,S,D(N,M<=500,0<S,D<N),接下来M行输入一条边的起点,终点,通过时间和通过花费.求花费最小的最短路,输入这条路径包含起点终点,通过时间和通过花费. trick: 找了半小时bug终于发现是给dis数组赋初值时范围是1~N,这道题点是从0~N-1的,故初次只通过了第0组数据,初始化改一下边界即可AC. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h>…

1030 Travel Plan (30)(30 分) A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her startin…

Source: PAT A1030 Travel Plan (30 分) Description: A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path b…

1030 Travel Plan (30 分) A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting ci…

Jimmy’s travel plan Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 341    Accepted Submission(s): 58 Problem Description Jimmy lives in a huge kingdom which contains lots of beautiful cities.…

#include<bits/stdc++.h> using namespace std; ; const int inf=0x3f3f3f3f; int mp[N][N]; bool vis[N]; int dis[N]; int n,m,s,D; int cost[N][N]; vector<int>path[N]; void Dijkstra() { fill(vis,vis+N,false); fill(dis,dis+N,inf); ;i<n;i++) dis[i]=…

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination.…

https://pintia.cn/problem-sets/994805342720868352/problems/994805464397627392 A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler t…

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination.…

题面 100 注意到ban的只会是一个子树,所以我们把原树转化为dfs序列. 然后题目就转化为,询问一段ban的区间,之后的背包问题. 比赛的时候,我想到这里,于是就开始想区间合并,于是搞了线段树合并,遂无果,爆零. 由于ban的是一段区间,所以肯定是将前缀和后缀合并. 我们预处理出前缀背包,和后缀背包. 然后合并两个背包就可以了. 具体的合并,Two pointers. 还要卡常. Code #include<iostream> #include<cstdio> #include…

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination.…

和1018思路如出一辙,先求最短路径,再dfs遍历 #include <iostream> #include <cstdio> #include <vector> #include<algorithm> using namespace std; ][], c[][];// distence cost vector<];//到达当前节点的前一个节点 vector<vector<int> > st; //最短距离的路径 vector…

#include <cstdio> #include <cstdlib> #include <vector> #include <queue> #include <unordered_map> #include <algorithm> #include <climits> #define CMB(ID1, ID2) (((ID1)<<9) | (ID2)) using namespace std; class…

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

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…

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…

早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1021. Deepest Root (25) 题意 无环连通图也可以视为一棵树,选定图中任意一点作为根,如果这时候整个树的深度最大,则称其为 deepest root. 给定一个图,按升序输出所有 deepest root.如果给定的图有多个连通分量,则输出连通分量的数量. 分析…