POJ 1573 Robot Motion】的更多相关文章

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: 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…
题目代号:POJ 1573 题目链接:http://poj.org/problem?id=1573 Language: Default Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14195   Accepted: 6827 Description A robot has been programmed to follow the instructions in its path. Instr…
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…
                                                                                                               Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11462   Accepted: 5558 Description A robot has been programmed to…
#define ONLINE_JUDGE #include<cstdio> #include <cstring> #include <algorithm> using namespace std; int A,B,sx,sy; char maz[101][101]; int vis[101][101]; const int dx[4]={0,1,0,-1}; const int dy[4]={-1,0,1,0}; int dir(char ch){ if(ch=='N'…
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…
又是被自己的方向搞混了 题意:走出去和遇到之前走过的就输出. #include <cstdlib> #include <iostream> #include<cstdio> using namespace std; #define N 110 int map[N][N],visit[N][N],n,m,flag;//n为x轴 m为y轴 int dir[][2]={{0,1},{1,0},{0,-1},{-1,0}};//e,s,w,n int setdire(char…
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…
-->Robot Motion 直接中文 Descriptions: 样例1 样例2 有一个N*M的区域,机器人从第一行的第几列进入,该区域全部由'N' , 'S' , 'W' , 'E' ,走到某个区域的时候只能按照该区域指定的方向进行下一步,问你机器人能否走出该片区域,若不能,输入开始绕圈的步数和圈的大小.操作指令如下: N 向上 S 向下 E 向右 W 向左 例如,假设机器人从网格1的北(顶)侧开始,从南(下)开始.机器人所遵循的路径如图所示.在离开网格之前,机器人在网格中执行10条指令.…