Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried despe…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 148156    Accepted Submission(s): 39500 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

传送门: HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1010 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1110 题目大意: 一只狗陷入了迷宫,他每秒必须走一步,并且不能重复走已经走过的地方,给定一个出口,要求在恰好为T的时间到达,如果可以,输出YES否则NO 不断的TLE,然后搜了下题解. 这题最漂亮的剪枝就在于奇偶的判断上.Orz DFS过程中,设终点的坐标为…

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 107138    Accepted Submission(s): 29131 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…

Sum It Up Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…

题目链接:http://ac.jobdu.com/problem.php?pid=1461 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1461 Tempter of the bone.cpp // Jobdu // // Created by PengFei_Zheng on 24/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. //…

求最久时间即在无环有向图里求最远路径 dfs+剪枝优化 从0节点(自己添加的)出发,0到1~n个节点之间的距离为1.mt[i]表示从0点到第i个节点眼下所得的最长路径 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<vector> using namespace std; const…

题目链接:  http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目大意:给定起点和终点,问刚好在t步时能否到达终点. 解题思路: 4个剪枝. ①dep>t剪枝 ②搜到一个解后剪枝 ③当前走到终点最少步数>满足条件还需要走的步数剪枝(关键) ③奇偶剪枝(关键):当前走到终点步数的奇偶性应该与满足条件还需要走的步数奇偶性一致. 其中三四两步放在一步中写:remain=abs(x-ex)+abs(y-ey)-abs(dep-t) 奇偶剪枝的原理:abs(…