E - A strange lift 【数值型BFS+上下方向】

【E - A strange lift 【数值型BFS+上下方向】】的更多相关文章

E - A strange lift 【数值型BFS+上下方向】

There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up K…

hdu 1548 A strange lift 宽搜bfs+优先队列

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if yo…

HDU 1548 A strange lift(最短路&&bfs）

A strange lift Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26943 Accepted Submission(s): 9699 Problem Description There is a strange lift.The lift can stop can at every floor as you want,…

HDU 1548 A strange lift(Dijkstra,简单BFS)

题目大意: 电梯有两个选项向上或向下,每层楼有一个参数ki,代表电梯可以再该楼层的基础上向上或向下移动ki层,限制条件是向上不能超过楼层总数n,向下不能少于一.输入总层数n和当前所在层数以及目标层数,然后是n个数分别代表第i层的移动范围.输出最少移动次数,若不可达,输出-1. 解题思路: 1.用Dijkstra算法,首先构建邻接矩阵,注意在构造时,要考虑i-k[i]<1和i+k[i]>n,i代表当前所在层. #include<string.h> #include<stdio.…

HDU1548：A strange lift(Dijkstra或BFS)

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1548 题意:电梯每层有一个数,例如第n层有个数k, 那么这一层只能上k层或下k层,但是不能低于一层或高于n层, 给定起点与终点,要求出最少要按几次键我的思路:这题可以用bfs或者用最短路(dijsktra) bfs AC代码 #include<cstdio>#include<iostream>#include<cstring>#include<queue>us…

D - Interesting Calculator 【数值型BFS+优先队列】

There is an interesting calculator. It has 3 rows of buttons. Row 1: button 0, 1, 2, 3, ..., 9. Pressing each button appends that digit to the end of the display. Row 2: button +0, +1, +2, +3, ..., +9. Pressing each button adds that digit to the disp…