网易游戏2016实习生招聘笔试题目--推箱子

题目描述

解题思路

对于输入的状态，找出玩家和箱子的位置，并使用坐标记录其位置，然后对于终端输入的走法，依次判断player按该走法一步一步的移动之后是否能够成功。由于思路比较简单，实际内容就是如何让计算机模拟这个推箱子的过程。步骤也很简单，只有上下左右移动，但是需要注意的是每次移动之前，需要对于题目描述的集中状态和情况进行判断，保证每一种情形都考虑到而且不重复判断。另外还需要注意的一点是，每次移动完成之后，需要对棋盘以及玩家和箱子的位置和状态进行更新，以便于真正模拟了游戏的过程。

实现代码

#include<iostream>#include<string>using namespace std;char chessboard[100][100]; // 棋盘布局char chessboard_bkup[100][100]; // 棋盘布局char path[200][10000]; // 推动路径int path_count[10000];string results[10000]; // 记录每一个玩法的最终胜负状态int main(){int N =0,M = 0,S = 0;cin>>N>>M>>S;int player_m = 0,player_n = 0; // 玩家的坐标int player_m_bkup = 0,player_n_bkup = 0;int box_m = 0,box_n = 0; // 箱子的坐标int box_m_bkup = 0,box_n_bkup = 0;bool flag_1 = true,flag_2 = true;for(int m=0;m<M;m++){ // M行cin>>chessboard_bkup[m];for(int n=0;n<strlen(chessboard_bkup[m]);n++){if(flag_1 == true){if(chessboard_bkup[m][n] == '1'){ // 玩家的位置player_m_bkup = m;player_n_bkup = n;flag_1 = false;}}if(flag_2 == true){if(chessboard_bkup[m][n] == '3'){ // 箱子的位置box_m_bkup = m;box_n_bkup = n;flag_2 = false;}}}}for(int i=0;i<S;i++) // 读取走法需要的步数和具体走法cin>>path_count[i]>>path[i];bool flag = false;for(int i=0;i<S;i++){// 每一次移动之后都需要判断是否已经到了目标位置// 在箱子的每次移动的时候需要判断将要移动的位置是否是目标位置，如果不是，需要对棋盘布局进行修改// 修改的棋盘是输入棋盘的一个副本flag = false;player_m = player_m_bkup;player_n = player_n_bkup;box_m = box_m_bkup;box_n = box_n_bkup;for(int p=0;p<M;p++)for(int q=0;q<N;q++)chessboard[p][q] = chessboard_bkup[p][q];for(int j=0;j<path_count[i];j++){if(path[i][j] == 'u'){ // 上移if(chessboard[player_m-1][player_n] == '3'){ // 移动方向上有箱子if(chessboard[player_m-2][player_n] == '0' || chessboard[player_m-2][player_n] == '2'){ // 移动方向没有被墙壁挡住if(chessboard[player_m-2][player_n] == '2'){chessboard[player_m][player_n] = '0';chessboard[player_m-1][player_n] = '1';player_m-=1;box_m-=1; flag = true;results[i] = "YES";}else{chessboard[player_m][player_n] = '0';chessboard[player_m-1][player_n] = '1';chessboard[player_m-2][player_n] = '3';player_m-=1;box_m-=1; }} // 若被挡住则什么也不做}else{if(chessboard[player_m-1][player_n] == '0' || chessboard[player_m-1][player_n] == '2'){if(chessboard[player_m-1][player_n] == '0'){chessboard[player_m][player_n] = '0';chessboard[player_m-1][player_n] = '1';player_m-=1; // 如果移动方向没有箱子，且没有墙壁阻挡，只移动玩家}else{player_m-=1;}}}}if(path[i][j] == 'd'){ // 下移if(chessboard[player_m+1][player_n] == '3' || chessboard[player_m+2][player_n] == '2'){ if(chessboard[player_m+2][player_n] == '2'){chessboard[player_m][player_n] = '0';chessboard[player_m+1][player_n] = '1';player_m+=1;box_m+=1; flag = true;results[i] = "YES";}else{chessboard[player_m][player_n] = '0';chessboard[player_m+1][player_n] = '1';chessboard[player_m+2][player_n] = '3';player_m+=1;box_m+=1;  } }else{if(chessboard[player_m+1][player_n] == '0' || chessboard[player_m+1][player_n] == '2'){if(chessboard[player_m][player_n+1] == '0'){chessboard[player_m][player_n] = '0';chessboard[player_m+1][player_n] = '1';player_m+=1;}else{player_m+=1;}}}}if(path[i][j] == 'l'){ // 左移if(chessboard[player_m][player_n-1] == '3'){ if(chessboard[player_m][player_n-2] == '0' || chessboard[player_m][player_n-2] == '2'){ if(chessboard[player_m][player_n-2] == '2'){chessboard[player_m][player_n] = '0';chessboard[player_m][player_n-1] = '1';player_n-=1;box_n-=1; flag = true;results[i] = "YES";}else{chessboard[player_m][player_n] = '0';chessboard[player_m][player_n-1] = '1';chessboard[player_m][player_n-2] = '3';player_n-=1;box_n-=1;  } }}else{if(chessboard[player_m][player_n-1] == '0' || chessboard[player_m][player_n-1] == '2'){if(chessboard[player_m][player_n+1] == '0'){chessboard[player_m][player_n] = '0';chessboard[player_m][player_n-1] = '1';player_n-=1;}else{player_n-=1;}}}}if(path[i][j] == 'r'){ // 右移if(chessboard[player_m][player_n+1] == '3'){ if(chessboard[player_m][player_n+2] == '0' || chessboard[player_m][player_n+2] == '2'){ if(chessboard[player_m][player_n+2] == '2'){chessboard[player_m][player_n] = '0';chessboard[player_m][player_n+1] = '1';player_n+=1;box_n+=1; flag = true;results[i] = "YES";}else{chessboard[player_m][player_n] = '0';chessboard[player_m][player_n+1] = '1';chessboard[player_m][player_n+2] = '3';player_n+=1;box_n+=1;  } }}else{if(chessboard[player_m][player_n+1] == '0' || chessboard[player_m][player_n+1] == '2'){if( chessboard[player_m][player_n+1] == '0'){chessboard[player_m][player_n] = '0';chessboard[player_m][player_n+1] = '1';player_n+=1;}else{player_n+=1;}}}}}if(flag == false)results[i] = "NO";}       for(int i=0;i<S;i++){cout<<results[i]<<endl;}return 0;}

测试输出

测试用例和测试结果如下：

箱子在目标点上输出为YES，否则输出为NO

代码中对于状态和棋盘都是使用的备份内容中实现的，由于时间紧迫，当时采用了这种临时的做法，后来看看，没有太多的必要使用一份备份来进行操作。不过对于推箱子的过程以及状态判断和棋盘更新部分的代码思路还是正确的。

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