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…

#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…

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…

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…

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…

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…

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…

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>…

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…

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…

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<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.…

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.…

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.…

(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…

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…

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.…

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.…

//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…

给定一个正整数数列,和正整数p,设这个数列中的最大值是M,最小值是m,如果M <= m * p,则称这个数列是完美数列. 现在给定参数p和一些正整数,请你从中选择尽可能多的数构成一个完美数列. 输入格式: 输入第一行给出两个正整数N和p,其中N(<= 105)是输入的正整数的个数,p(<= 109)是给定的参数.第二行给出N个正整数,每个数不超过109. 输出格式: 在一行中输出最多可以选择多少个数可以用它们组成一个完美数列. 输入样例: 10 8 2 3 20 4 5 1 6 7 8…

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…

给定一个正整数数列,和正整数p,设这个数列中的最大值是M,最小值是m,如果M <= m * p,则称这个数列是完美数列. 现在给定参数p和一些正整数,请你从中选择尽可能多的数构成一个完美数列. 输入格式: 输入第一行给出两个正整数N和p,其中N(<= 105)是输入的正整数的个数,p(<= 109)是给定的参数.第二行给出N个正整数,每个数不超过109. 输出格式: 在一行中输出最多可以选择多少个数可以用它们组成一个完美数列. 输入样例: 10 8 2 3 20 4 5 1 6 7 8…

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/994805260223102976/problems/994805291311284224 给定一个正整数数列,和正整数 p,设这个数列中的最大值是 M,最小值是 m,如果 M≤mp,则称这个数列是完美数列. 现在给定参数 p 和一些正整数,请你从中选择尽可能多的数构成一个完美数列. 输入格式: 输入第一行给出两个正整数 N 和 p,其中 N(≤)是输入的正整数的个数,p(≤)是给定的参数.第二行给出 N个正整数,每个数不超过 1.…

#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…

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/=; } ;…