HDU 1072 Nightmare】的更多相关文章

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.…
http://acm.hdu.edu.cn/showproblem.php?pid=1072 遇到Bomb-Reset-Equipment的时候除了时间恢复之外,必须把这个点做标记不能再走,不然可能造成死循环,也得不到最优解. #include <cstdio> #include <cstring> #include <queue> using namespace std; struct point { int x,y,time,step; }; point s; in…
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…
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…
Nightmare Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14260    Accepted Submission(s): 6930 Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth with a ti…
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 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 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…
之前的BFS都是需要一个标记数组,但这个题不一样,因为可能一个格子不止走一次. 那么我们就要寻找新的入队条件:left比上次经过的时候大才入队(left表示上次经过该点时剩余的时间). 为什么呢?我们重复走过一个点只有一个可能,那就是为了去取那个,所以如果取完后再回头经过这个点的时候剩余时间变多了,我们的目的就达到了. left数组初值为0 优化: 重置时间的装置最多取一次就够了,所以可以再开一个标记数组vis记录装置是否用过. //#define LOCAL #include <cstdio>…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1072 题目大意:走迷宫.走到装置点重置时间,到达任一点时的时间不能为0,可以走重复路,求出迷宫最短时间. 解题思路: vis的第三维标记一下到这个格子的时间. 尽管可以格子可以重复走,但在相同时间到这个格子是没有意义的. 小心一下时间不能为0的问题就行了. #include "cstdio" #include "queue" #include "cstrin…