[hdu P3085] Nightmare Ⅱ Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2885 Accepted Submission(s): 806 Problem DescriptionLast night, little erriyue had a horrible nightmare. He dreamed that he a…

HDU 3085 Nightmare Ⅱ(噩梦 Ⅱ) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)   Problem Description - 题目描述 Last night, little erriyue had a horrible nightmare. He dreamed that he and his girl friend were trapped in a…

HDU - 3085 Nightmare Ⅱ 双向BFS,建立两个队列,让男孩女孩一起走 鬼的位置用曼哈顿距离判断一下,如果该位置与鬼的曼哈顿距离小于等于当前轮数的两倍,则已经被鬼覆盖 #include <cstdio> #include <queue> #include <algorithm> #include <cmath> #include <cctype> #include <cstring> using namespace…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1072 Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes.…

Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6…

http://acm.hdu.edu.cn/showproblem.php?pid=3085 出的很好的双向bfs,卡时间,普通的bfs会超时 题意方面: 1. 可停留 2. ghost无视墙壁 3. 需要检查两次某个地点是否有ghost,正要到达的时候(t),以及即将启程的时候(t+1). 在编程时需要注意的是: 当两个人汇合时,不需要检查即将启程的那次. M可以走3步,在这3步中间只需要检查是否能到达(t) 出问题在: 1. 检查时间处理的不清晰 2.普通bfs会超时 3.双向bfs需要完全…

http://acm.hdu.edu.cn/showproblem.php?pid=1240 开始没仔细看题,看懂了发现就是一个裸的bfs,注意坐标是三维的,然后每次可以扩展出6个方向. 第一维代表在第几层.后两维代表行和列. #include <cstdio> #include <cstring> #include <queue> using namespace std; struct point { int x,y,z,step; bool operator <…

Nightmare Ⅱ Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 886    Accepted Submission(s): 185 Problem Description Last night, little erriyue had a horrible nightmare. He dreamed that he and hi…

题目链接 Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bom…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3085 题目大意:给你一张n*m地图上,上面有有 ‘. ’:路 ‘X':墙 ’Z':鬼,每秒移动2步,可以穿墙,开始有两个,每开始时鬼先动. ‘M’:男生,每秒可走3步. ‘G’:女生,每秒可走1步. 解题思路:第一次写双向BFS,写了我好久,开始还是想着先bfs计算step[x][y][t]把每个位置被鬼占据的时间处理一下然后再用双向BFS计算两人相遇时间.后来发现因为鬼可以穿墙,可以直接用曼哈顿距…