Tempter of the Bone [从零开始DFS(1)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] -小…

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): 82702    Accepted Submission(s): 22531 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

http://acm.hdu.edu.cn/showproblem.php?pid=1010   //题目链接 http://ycool.com/post/ymsvd2s//一个很好理解剪枝思想的博客 http://blog.csdn.net/chyshnu/article/details/6171758//一个很好举例的博客 Problem Description The doggie found a bone in an ancient maze, which fascinated him…

  如果所给的时间(步数) t 小于最短步数path,那么一定走不到. 若满足t>path.但是如果能在恰好 t 步的时候,走到出口处.那么(t-path)必须是二的倍数. 关于第二种方案的解释: 这种方案学名为“奇偶剪枝”.我们已知了最短的步数就是直角三角形的两条直角边,实际上的路径却不一定非要沿着这两条边走的.仔细看看只要是移动方向一直是右.下,那么走到的时候总步数也一定是path的.然而由于墙的存在或许我们不可能一直右.下的走下去.为了避开墙,我们可能会向左走,向上走等等.但为了到达目的地…

题目大意:输入三个整数N,M,T.在接下来的N行.M列会有一系列的字符.其中S表示起点,D表示终点. .表示路 . X表示墙...问狗能有在T秒时到达D.如果能输出YES, 否则输出NO 解题思路:DFS+剪枝 代码如下: /* * 1010_3.cpp * * Created on: 2013年8月16日 * Author: Administrator */ #include <iostream> using namespace std; bool flag; int N, M, T; in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目大意: 输入 n m t,生成 n*m 矩阵,矩阵元素由 ‘.’ 'S' 'D' 'X' 四类元素组成. S'代表是开始位置: 'D'表示结束位置:'.'表示可以走的路:'X'表示是墙. 问:从‘S’  能否在第 t 步 正好走到 'D'. 解题思路: 平常心态做dfs即可,稍微加个奇偶剪枝,第一次做没经验,做过一次下次就知道怎么做了.最后有代码注释解析. AC Code: #includ…

http://acm.hdu.edu.cn/showproblem.php?pid=1010 这题就是问能不能在t时刻走到门口,不能用bfs的原因大概是可能不一定是最短路路径吧. 但是这题要过除了细心外,还需要强力的剪枝. 奇偶性剪枝:参考 http://www.cppblog.com/Geek/archive/2010/04/26/113615.html #include <iostream> #include <cstring> #include <cstdio>…

需要剪枝否则会超时,然后就是基本的深搜了 #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<queue> #define INF 0x3f3f3f3…

<题目链接> 题目大意:一个迷宫,给定一个起点和终点,以及一些障碍物,所有的点走过一次后就不能再走(该点会下陷).现在问你,是否能从起点在时间恰好为t的时候走到终点. 解题分析:本题恰好要在某一时刻到达,所以需要用到可行性剪枝中的奇偶剪枝,如果在某一点,它所剩的步数与到终点的最短距离之差是偶数,说明这种情况有可能恰好在规定时刻到达终点,否则不可能,将其剪去. #include <bits/stdc++.h> using namespace std; ][]; ][]; int n,…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1010 Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 131057    Accepted Submission(s): 35308 Problem Description The doggie fou…

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): 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): 88774    Accepted Submission(s): 24159 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) 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…

M - Tempter of the Bone Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u 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 d…

Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 149833    Accepted Submission(s): 39945 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

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…

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 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…

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…

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…

题意: 必须在第t秒走到格子D上,S为起点,D为终点,点就是可以走,X就是墙. 思路: 将迷宫外围四面都筑墙‘X’.深度搜索+奇偶剪枝,再加一个剪枝“无法在指定时间内到达”. #include <iostream> #include <vector> #include <string> #include <math.h> using namespace std; vector<string> v; int n,m; int x_1,y_1,x_2…

http://acm.hdu.edu.cn/showproblem.php?pid=1010 题意:就是给出了一个迷宫,小狗必须经过指定的步数到达出口,并且每个格子只能走一次. 首先先来介绍一下奇偶性剪枝: 在这道题目中,如果使用剪枝的话,可以节省不少的时间. 在这道题目中,每次dfs循环时都可以判断一下小狗当前位置与终点所相差的步数,如果不为偶数的话,说明到达不了终点,就可以退出这个循环,不必继续dfs了. 在这道题目中,由于每个格子只能经过一次,所以经过一次后,可以把该点位置改为'X',然后…

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…

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…