题意:从S走到D,能不能恰好用T时间. 析:这个题时间是恰好,并不是少于T,所以用DFS来做,然后要剪枝,不然会TEL,我们这样剪枝,假设我们在(x,y),终点是(ex,ey), 那么从(x, y)到(ex, ey),要么时间正好是T-你已经走过的时间,要么要向别的地方先拐一下,以凑出这个正好时间,既然要拐一下,那么一定要回来, 所以时间肯定得是偶数,要不然完不成(回不来), 所以(t - abs(ex-x) - abs(ey-y) - cnt ),如果是奇数就剪枝.然而用C++交就TLE,用G

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

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): 92175    Accepted Submission(s): 25051 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): 135529    Accepted Submission(s): 36393 Problem Description The doggie found a bone in an ancient maze, which fascinated him

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 146511    Accepted Submission(s): 39059 Problem Description The doggie found a bone in an ancient maze, which fascinated him

Tempter of the Bone Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 5   Accepted Submission(s) : 1 Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. Howe

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) 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

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 90328    Accepted Submission(s): 24554 Problem Description The doggie found a bone in an ancient maze, which fascinated him a

Tempter of the Bone 直接上中文了 Descriptions: 暑假的时候,小明和朋友去迷宫中寻宝.然而,当他拿到宝贝时,迷宫开始剧烈震动,他感到地面正在下沉,他们意识到这是一个陷阱!他们想尽一切办法逃出去.迷宫是一个大小为 N*M 的长方形,迷宫中有一扇门.一开始,门是关着的,他会在第 t 秒的时间打开.因为,小明和朋友必须在第 t 秒到大门口.每一秒,他都可以向上下左右四个方向移动一个点.一旦他移动了,他刚才所在的点就消失,(这意味着他不能回到他已经走过的点).他不能在一个

Tempter of the Bone Time Limit: 2 Seconds      Memory Limit: 65536 KB 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 rea

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): 70032    Accepted Submission(s): 19286 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): 88774    Accepted Submission(s): 24159 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

传送门:Tempter of the Bone 大意是给一个矩阵,叫你是否可以在给定的可走路径上不重复地走,在最后一秒走到终点. 我用了两个剪枝,且称其为简直001和剪枝002,事实证明001不要都可以,002不要也能过--||.就当练习一下剪枝. 特别是002很有用: if( d % 2 != (Time-t) % 2 ) return false ; (d=~x+~y) #include<cstdio> #include<cstdlib> #include<iostrea

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

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 102071    Accepted Submission(s): 27649 Problem Description The doggie found a bone in an ancient maze, which fascinated him

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. //