PAT_A1030#Travel Plan】的更多相关文章

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…
(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…
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…
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…
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…
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…
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.…
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…
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…