解题心得: 1.注意审题,此题是在规定的时间达到规定的地点,不能早到也不能晚到.并不是最简单的dfs 2.在规定时间达到规定的地点有几个剪枝: 一.公式:所需的步骤 - x相差行 - y相差列 = 偶数.(这个解释很简单,无论怎么走,都可以把走的总路程分解到x方向和y方向,哪怕反向走,走回来后的步骤+反向走的步骤也一定是偶数,假设运用正交分解走到了终点(上面公式),但是步骤没走够,可以以最终的目标点为起点任选两格来回走消耗步骤,但是减去正交分解的步数后若是奇数,那么在终点来回走将步数消耗完之后必…

http://acm.hdu.edu.cn/showproblem.php?pid=1010 用到了奇偶剪枝: 0 1 0 1 1 0 1 0          如图,设起点为s,终点为e,s->e的最短步数为t.   sum=t+extra. [因为extra无论怎么走,最终都要返回其中一条最短路径上,即有来回,所以extra=2*某个数,一定是偶数. 0 1 0 1 1 0 1 0 [所以此题中,v-step表示还可以走多少步,abs(x-x2)+abs(y-y2)为此刻点到终点位置,由上面…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 82702    Accepted Submission(s): 22531 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 58766    Accepted Submission(s): 15983 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

HDU 1010 题目大意:给定你起点S,和终点D,X为墙不可走,问你是否能在 T 时刻恰好到达终点D. 参考: 奇偶剪枝 奇偶剪枝简单解释: 在一个只能往X.Y方向走的方格上,从起点到终点的最短步数为T1,并记其他任意走法所需步数为T2,则T2-T1一定为偶数. 即若某一点到终点的最短步数为T1,且T3-T1为奇数,则一定无法话费T3步恰好到达终点. /*HDU 1010 ------ Tempter of the Bone DFS*/ #include <cstdio> #include…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 89317    Accepted Submission(s): 24279 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 94669    Accepted Submission(s): 25669 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 92175    Accepted Submission(s): 25051 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

题意  一仅仅狗要逃离迷宫  能够往上下左右4个方向走  每走一步耗时1s  每一个格子仅仅能走一次且迷宫的门仅仅在t时刻打开一次  问狗是否有可能逃离这个迷宫 直接DFS  直道找到满足条件的路径  或者走全然部可能路径都不满足 注意剪枝  当前位置为(r,c)  终点为(ex,ey) 剩下的时间为lt  当前点到终点的直接距离为  d=(ex-r)+(ey-c)   若多走的时间rt=lt-d<0 或为奇数时  肯定是不可能的  能够自己在纸上画一下 每一个点仅仅能走一次的图  走弯路的话多…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 125945    Accepted Submission(s): 33969 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…