Asteroids! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3159    Accepted Submission(s): 2106 Problem Description You're in space.You want to get home.There are asteroids.You don't want to hit…

Asteroids! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2599    Accepted Submission(s): 1745 Problem Description You're in space. You want to get home. There are asteroids. You don't want to…

题目链接 Problem Description You're in space.You want to get home.There are asteroids.You don't want to hit them.   Input Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the…

Asteroids! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6174    Accepted Submission(s): 3876 Problem Description You're in space.You want to get home.There are asteroids.You don't want to hit…

//这道题做完我只有 三个感受  第一:坑: 第二 : 坑! 第三:还是坑! 咳咳  言归正传  WA了无数次之后才发现是输入进去时坐标时z, y, x的顺序输入的 题解   :  类似胜利大逃亡 只不过给你了起始坐标和终点坐标, 让你输出到达目标点所需最少步数: 输出时第一个输出时是START读入的map大小值n;第二个是所求步数 //细节: 1.读入:   读入时分别两次%S把没用的START 和 END读取掉: 2.输出时输出 三维坐标大小值n, 以及步数: 3.输入进去时开始点和结束点坐…

三维广搜 #include <cstdio> #include <iostream> #include <cstring> #include <queue> using namespace std; struct node { int x,y,z; int steps; }start,end,next; ]={,,,,,-}; ]={,,-,,,}; ]={,-,,,,}; ][][]; int n,res; bool check(node &a)…

普通的三维广搜,须要注意的是输入:列,行,层 #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> #define M 11 using namespace std; int dir[6][3]={{0,1,0},{0,-1,0},{1,0,0},{-1,0,0},{0,0,1},{0,0,-1}};//6个方向 int…

题目链接:点击链接 简单BFS,和二维的做法相同(需注意坐标) 题目大意:三维的空间里,给出起点和终点,“O”表示能走,“X”表示不能走,计算最少的步数 #include <iostream> #include <stdio.h> #include <string.h> #include <queue> using namespace std; char map[11][11][11]; int v[11][11][11],d[6][3] = { {1,0,…

题意:给出一个三维的空间,给出起点和终点,问是否能够到达终点 和上一题一样,只不过这一题的坐标是zxy输入的, 因为题目中说的是接下来的n行中分别是由n*n的矩形组成的,所以第一个n该是Z坐标,n*n的矩形为底面,为x,y坐标 -----还是注意输入的方式--- #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #de…

给出一个三维的迷宫以及起点和终点,求能否到大终点,若果能输出最短步数 三维的问题无非就是变成了6个搜索方向 最后强调一下xyz的顺序,从输入数据来看,读入的顺序是map[z][x][y] 总之,这是很基础的一道题 //#define LOCAL #include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <algorithm> using n…