POJ 3635 优先队列BFS】的更多相关文章

(感谢lyd学长的幻灯片) 注意vis数组的应用 在vis[i][j]中 i表示到了第i个点 j表示还剩j升油 vis[i][j]表示最小话费. 这样只需搜到话费比它少的更新入堆就OK了 //By: Sirius_Ren #include <queue> #include <cstdio> #include <cstring> using namespace std; struct node{int num,wei,oil;}jy,temp; int next[2000…
After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited. Maybe you could have saved some money if you were a bit more clever about where you filled your fu…
题目链接:http://poj.org/problem?id=2449 Time Limit: 4000MS Memory Limit: 65536K Description "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a story. &quo…
[题目链接] http://poj.org/problem?id=3635 [算法] 优先队列BFS 实现类似于堆优化dijkstra [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include &…
链接: http://acm.hdu.edu.cn/showproblem.php?pid=1026 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29096#problem/D Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(…
题目链接:http://poj.org/problem?id=3635 题意题解等均参考:POJ 3635 - Full Tank? - [最短路变形][优先队列优化Dijkstra]. 一些口胡: 说实话,上次写类似的二维状态最短路Gym 101873C - Joyride - [最短路变形][优先队列优化Dijkstra],我没能把手写二叉堆优化Dijkstra的给写出来. 这次费了点功夫,也算是给写出来了,需要注意的点还是有点多的.而且我终于深刻理解为啥不推荐手写二叉堆了,主要是代码量相比…
任意门:http://poj.org/problem?id=3635 Full Tank? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8388   Accepted: 2734 Description After going through the receipts from your car trip through Europe this summer, you realised that the gas pri…
Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 521    Accepted Submission(s): 217   Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…
普通BFS:每个状态只访问一次,第一次入队时即为该状态对应的最优解. 优先队列BFS:每个状态可能被更新多次,入队多次,但是只会扩展一次,每次出队时即为改状态对应的最优解. 且对于优先队列BFS来说,每次存入队列的不光是像普通BFS的状态,还有当前状态对应的代价,并且是依据最小代价进行扩展.每次状态被更近之后,将其入队. 对于本题来说 状态选取:当前所在城市,当前油量 代价函数:当前状态所对应的最小花费 代码如下: #include <cstdio> #include <iostream…
题目链接:http://codeforces.com/problemset/problem/677/D 题意: 有 $n \times m$ 的网格,每个网格上有一个棋子,棋子种类为 $t[i][j]$,棋子的种类数为 $p$. 现在出发点为 $(1,1)$,必须按照种类 $1 \sim p$ 进行移动,即从种类 $x$ 的棋子出发,下一个目标必须是 $x+1$ 才行,直到走到种类为 $p$ 的棋子就终止.求最短路径. 题解: 我们先把棋子按照种类分组,分成 $p$ 组. $dp[i][j]$…