HDOJ(HDU).1035 Robot Motion [从零开始DFS(4)] 点我挑战题目 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)…

Problem Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are N north (up the page)S south (down the page)E east…

题目大意:输入三个整数n,m,k,分别表示在接下来有一个n行m列的地图.一个机器人从第一行的第k列进入.问机器人经过多少步才能出来.如果出现了循环 则输出循环的步数 解题思路:DFS 代码如下(有详细的解释): /* * 1035_1.cpp * * Created on: 2013年8月17日 * Author: Administrator */ /*简单搜索题,看注释: */ #include<iostream> #include<cstdio> #include<cst…

Robot Motion Problem Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are  N north (up the page) S south (down…

虽然做出来了,还是很失望的!!! 加油!!!还是慢慢来吧!!! >>>>>>>>>>>>>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<<<<<<< >>&g…

嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1035 这道题比较简单,但自己一直被卡,原因就是在读入mp这张字符图的时候用了scanf被卡... 注意初始化和dfs边界:如果超出图或者曾经被标记过则输出 AC代码: #include<cstdio> #include<cstring> #include<iostream> using namespace std; ][]; ][]; inline void df…

Problem Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are N north (up the page) S south (down the page) E ea…

Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7974    Accepted Submission(s): 3685 Problem Description A robot has been programmed to follow the instructions in its path. Instruct…

DFS and  BFS 在解题前我们还是大致讲一下dfs与bfs的.(我感觉我不会bfs) 1.DFS dfs(深度优先算法) 正如其名,dfs是相当的深度,不走到最深处绝不回头的那种. 深度优先搜索是一种枚举所有完整路径以遍历所有情况的搜索方法. 而使用递归可以很好地实现深度优先搜索. 在使用递归时,系统会调用一个叫系统栈的东西来存放递归中每一层的状态,因此使用递归来实现DFS的本质其实还是栈. 下面DFS的模版: void dfs() { //参数用来表示状态 if(到达终点状态) { .…

题目在这里 题意 : 问你按照图中所给的提示走,多少步能走出来??? 其实只要根据这个提示走下去就行了.模拟每一步就OK,因为下一步的操作和上一步一样,所以简单dfs.如果出现loop状态,只要记忆每个所到的点的第一次的步数,最后总步数减掉它即可 /************************************************************************* > File Name: poj1573.cpp > Author: YeGuoSheng >…

原题大意:原题链接 给出一个矩阵(矩阵中的元素均为方向英文字母),和人的初始位置,问是否能根据这些英文字母走出矩阵.(因为有可能形成环而走不出去) 此题虽然属于水题,但是完全独立完成而且直接1A还是很开心的 注意:对于形成环的情况则从进入环的交点处重新走一遍,记录步数即可 #include<cstdio> #include<cstring> int n,m,p; ][]; ][]; bool can(int x,int y) { ||x>n||y<||y>m||v…

http://acm.hdu.edu.cn/showproblem.php?pid=1035 Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5591    Accepted Submission(s): 2604 Problem Description A robot has been programmed t…

Problem Description People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ...…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2069 Problem Description Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.For example, if we have 11…

Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says."The second problem is, given an positive integer N, we define an equation like this:  N=a[1]+a[2]+a[3]+...+a[m…

Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12507   Accepted: 6070 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in…

2015浙江财经大学ACM有奖周赛(一) 题解报告 命题:丽丽&&黑鸡 这是命题者原话. 题目涉及的知识面比较广泛,有深度优先搜索.广度优先搜索.数学题.几何题.贪心算法.枚举.二进制等等... 有些题目还需要大家对程序的效率做出优化..大一的小宝宝可能有一些吃不消..当成是一种体验就好了. 题解目录: ZUFE OJ 2307: 最长连续不下降子串 ZUFE OJ 2308: Lucky Number ZUFE OJ 2309: 小明爱吃面 ZUFE OJ 2310: 小明爱消除 ZUF…

Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12856   Accepted: 6240 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in…

Abandoned country Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4477    Accepted Submission(s): 1124 Problem Description An abandoned country has n(n≤100000) villages which are numbered from 1…

OwO 题目含义都是一样的,只是数据范围扩大了 对于n<=7的问题,我们直接暴力搜索就可以了 对于n<=1000的问题,我们不难联想到<主旋律>这一道题 没错,只需要把方程改一改就可以了 首先我们考虑不合法的方案强连通分量缩点后一定是DAG 考虑子问题:DAG计数 做法可以参考<cojs DAG计数1-4 题解报告> 这里给出转移方程 f(n)=sigma((-1)^(k-1)*C(n,k)*2^(k*(n-k))*f(n-k)) 如果考虑上强连通分量缩点的情况呢? 我…

Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8180    Accepted Submission(s): 3771 Problem Description A robot has been programmed to follow the instructions in its path. Instruc…

Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10072    Accepted Submission(s): 4697 Problem Description A robot has been programmed to follow the instructions in its path. Instruc…

C - Catch That Cow POJ - 3278 我心态崩了,现在来回顾很早之前写的简单搜索,好难啊,我怎么写不出来. 我开始把这个写成了dfs,还写搓了... 慢慢来吧. 这个题目很明显是一个很简单的搜索题,但是如果用dfs的话很容易出现问题,而且复杂度不低. 所以选择用bfs. 这个知道用bfs就应该比较简单了. 就是有三种走的方式,因为k最大为100000  所以我们要限制一下,不然会超时. #include <cstdio> #include <cstdlib>…

空搜索: GET /_search hits: total 总数 hits 前10条数据 hits 数组中的每个结果都包含_index._type和文档的_id字段,被加入到_source字段中这意味着在搜索结果中我们将可以直接使用全部文档. 每个节点都有一个_score字段,这是相关性得分(relevance score),它衡量了文档与查询的匹配程度.默认的,返回的结果中关联性最大的文档排在首位:这意味着,它是按照_score降序排列的.没有指定任何查询,所以所有文档的相关性是一样的,因此所…

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=284 题意:在一个给定图中,铁墙,河流不可走,砖墙走的话,多花费时间1,问从起点到终点至少需要多少时间. 思路:简单广搜~ 代码如下: #include "stdio.h" //nyoj 284 坦克大战 简单搜索 #include "string.h" #include "queue" using namespace std; #def…

Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11065   Accepted: 5378 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in…

题目地址:http://poj.org/problem?id=1573 /* 题意:给定地图和起始位置,robot(上下左右)一步一步去走,问走出地图的步数 如果是死循环,输出走进死循环之前的步数和死循环的步数 模拟题:used记录走过的点,因为路线定死了,所以不是死循环的路只会走一次,可以区分出两个步数 注意:比较坑的是,如果不是死循环,临界(走进去就出来)步数是1:而死循环却是0. 这里WA几次... */ #include <cstdio> #include <iostream&g…

Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11262 Accepted: 5482 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a g…

Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12978   Accepted: 6290 Description A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in…

1.Link: http://poj.org/problem?id=1573 http://bailian.openjudge.cn/practice/1573/ 2.Content: Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10856   Accepted: 5260 Description A robot has been programmed to follow the instru…