Curling 2.0
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 14090   Accepted: 5887

Description

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game
is to lead the stone from the start to the goal with the minimum number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed
until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.



Fig. 1: Example of board (S: start, G: goal)

The movement of the stone obeys the following rules:

  • At the beginning, the stone stands still at the start square.
  • The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited.
  • When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)).
  • Once thrown, the stone keeps moving to the same direction until one of the following occurs:
    • The stone hits a block (Fig. 2(b), (c)).

      • The stone stops at the square next to the block it hit.
      • The block disappears.
    • The stone gets out of the board.
      • The game ends in failure.
    • The stone reaches the goal square.
      • The stone stops there and the game ends in success.
  • You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.



Fig. 2: Stone movements

Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.

With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).



Fig. 3: The solution for Fig. D-1 and the final board configuration

Input

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.

Each dataset is formatted as follows.

the width(=w) and the height(=h) of the board 

First row of the board
 

... 

h-th row of the board

The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.

Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.

0 vacant square
1 block
2 start position
3 goal position

The dataset for Fig. D-1 is as follows:

6 6 

1 0 0 2 1 0 

1 1 0 0 0 0 

0 0 0 0 0 3 

0 0 0 0 0 0 

1 0 0 0 0 1 

0 1 1 1 1 1

Output

For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.

Sample Input

2 1
3 2
6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1
6 1
1 1 2 1 1 3
6 1
1 0 2 1 1 3
12 1
2 0 1 1 1 1 1 1 1 1 1 3
13 1
2 0 1 1 1 1 1 1 1 1 1 1 3
0 0

Sample Output

1
4
-1
4
10
-1

冰球,我更喜欢把它叫推箱子。题意就是它这个箱子能够上下左右四个方向推。碰到边界就跑出去,碰到障碍物停止。之后碰到的障碍物就会在地图上消失。问从S開始经过几步能够达到E。超过10步的话就输出-1。

深搜,自己觉得还是比較简单。毕竟模拟一遍,也不须要弄什么算法。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; #define up 0
#define down 1
#define left 2
#define right 3 int value[25][25];
int row,col,flag,tui_x,tui_y,result,result_zuizhong; bool can_tui(int x_r,int y_r,int dir)
{
if(dir==up||dir==down)
{
int i,j;
if(dir==up)
{
if(value[x_r-1][y_r]==1)
return false;
for(i=1;i<x_r;i++)
{
if(value[i][y_r])
return true;
}
return false;
}
else
{
if(value[x_r+1][y_r]==1)
return false;
for(i=x_r+1;i<=row;i++)
{
if(value[i][y_r])
return true;
}
return false;
}
}
else
{
int i,j;
if(dir==left)
{
if(value[x_r][y_r-1]==1)
return false;
for(i=1;i<y_r;i++)
{
if(value[x_r][i])
return true;
}
return false;
}
else
{
if(value[x_r][y_r+1]==1)
return false;
for(i=y_r+1;i<=col;i++)
{
if(value[x_r][i])
return true;
}
return false;
}
}
} void tui(int x_r,int y_r,int dir)
{
int i,j;
if(dir==up)
{
for(i=x_r-1;i>=1;i--)
{
if(value[i][y_r]==1)
{
tui_x=i+1;
tui_y=y_r;
return;
}
if(value[i][y_r]==3)
{
tui_x=i;
tui_y=y_r;
return;
}
}
}
else if(dir==down)
{
for(i=x_r+1;i<=row;i++)
{
if(value[i][y_r]==1)
{
tui_x=i-1;
tui_y=y_r;
return;
}
if(value[i][y_r]==3)
{
tui_x=i;
tui_y=y_r;
return;
}
}
}
else if(dir==left)
{
for(i=y_r-1;i>=1;i--)
{
if(value[x_r][i]==1)
{
tui_x=x_r;
tui_y=i+1;
return;
}
if(value[x_r][i]==3)
{
tui_x=x_r;
tui_y=i;
return;
}
}
}
else
{
for(i=y_r+1;i<=col;i++)
{
if(value[x_r][i]==1)
{
tui_x=x_r;
tui_y=i-1;
return;
}
if(value[x_r][i]==3)
{
tui_x=x_r;
tui_y=i;
return;
}
}
}
} void dfs(int x_r,int y_r,int step,int dir)
{
if(step>11)
return;
if(value[x_r][y_r]==3)
{
flag=1;
result=min(result,step);
return;
}
if(dir==up)
{
value[x_r-1][y_r]=0;
}
else if(dir==down)
{
value[x_r+1][y_r]=0;
}
else if(dir==left)
{
value[x_r][y_r-1]=0;
}
else if(dir==right)
{
value[x_r][y_r+1]=0;
} if(can_tui(x_r,y_r,up))
{
tui(x_r,y_r,up);
int temp_x=tui_x;
int temp_y=tui_y;
dfs(temp_x,temp_y,step+1,up);
}
if(can_tui(x_r,y_r,left))
{
tui(x_r,y_r,left);
int temp_x=tui_x;
int temp_y=tui_y;
dfs(temp_x,temp_y,step+1,left);
}
if(can_tui(x_r,y_r,down))
{
tui(x_r,y_r,down);
int temp_x=tui_x;
int temp_y=tui_y;
dfs(temp_x,temp_y,step+1,down);
}
if(can_tui(x_r,y_r,right))
{
tui(x_r,y_r,right);
int temp_x=tui_x;
int temp_y=tui_y;
dfs(temp_x,temp_y,step+1,right);
} if(dir==up)
{
value[x_r-1][y_r]=1;
}
else if(dir==down)
{
value[x_r+1][y_r]=1;
}
else if(dir==left)
{
value[x_r][y_r-1]=1;
}
else if(dir==right)
{
value[x_r][y_r+1]=1;
}
} void solve()
{
int i,j;
for(i=row;i>=1;i--)
{
for(j=col;j>=1;j--)
{
if(value[i][j]==2)
{
value[i][j]=0;
result=11;
dfs(i,j,0,-1);
if(flag)
{
result_zuizhong =min(result,result_zuizhong);
}
return;
}
}
}
} int main()
{
int i,j;
while(cin>>col>>row)
{
if(col+row==0)
break;
flag=0;
result_zuizhong=11;
memset(value,0,sizeof(value)); for(i=1;i<=row;i++)
{
for(j=1;j<=col;j++)
{
cin>>value[i][j];
}
}
solve();
if(result_zuizhong==11)
cout<<-1<<endl;
else
cout<<result_zuizhong<<endl; }
return 0;
}

POJ 3009:Curling 2.0 推箱子的更多相关文章

  1. POJ 3009 Curling 2.0【带回溯DFS】

    POJ 3009 题意: 给出一个w*h的地图,其中0代表空地,1代表障碍物,2代表起点,3代表终点,每次行动可以走多个方格,每次只能向附近一格不是障碍物的方向行动,直到碰到障碍物才停下来,此时障碍物 ...

  2. poj 3009 Curling 2.0 (dfs )

    Curling 2.0 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11879   Accepted: 5028 Desc ...

  3. poj 3009 Curling 2.0

    题目来源:http://poj.org/problem?id=3009 一道深搜题目,与一般搜索不同的是,目标得一直往一个方向走,直到出界或者遇到阻碍才换方向. 1 #include<iostr ...

  4. POJ 3009 Curling 2.0(DFS + 模拟)

    题目链接:http://poj.org/problem?id=3009 题意: 题目很复杂,直接抽象化解释了.给你一个w * h的矩形格子,其中有包含一个数字“2”和一个数字“3”,剩下的格子由“0” ...

  5. POJ 3009 Curling 2.0 {深度优先搜索}

    原题 $On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules ...

  6. POJ 3009 Curling 2.0 回溯,dfs 难度:0

    http://poj.org/problem?id=3009 如果目前起点紧挨着终点,可以直接向终点滚(终点不算障碍) #include <cstdio> #include <cst ...

  7. poj 3009 Curling 2.0( dfs )

    题目:http://poj.org/problem?id=3009 参考博客:http://www.cnblogs.com/LK1994/ #include <iostream> #inc ...

  8. 【POJ】3009 Curling 2.0 ——DFS

    Curling 2.0 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11432   Accepted: 4831 Desc ...

  9. 【原创】poj ----- 3009 curling 2 解题报告

    题目地址: http://poj.org/problem?id=3009 题目内容: Curling 2.0 Time Limit: 1000MS   Memory Limit: 65536K Tot ...

随机推荐

  1. [Deepin 15] 编译安装 MySQL-5.6.35

    在 Ubuntu 下,先前一直是 二进制包解压安装,详情参考: http://www.cnblogs.com/52php/p/5680906.html 现改为 源码编译安装: #!/bin/bash ...

  2. solaris 软件包地址

      1. http://www.opencsw.org/ 在Solaris 10下 1.安装pkgutil pkgadd -d http://get.opencsw.org/now 2.查询有那些PK ...

  3. linux后台开发核心技术

    3. 常用STL的使用 3.1. string (1)string类的实现(使用strlen.strcpy.strcat.strcmp等,注意判NULL). (2)C++字符串和C字符串的转换:dat ...

  4. SpringBoot配置多数据源

    原文:https://www.jianshu.com/p/033e0ebeb617 项目中用到了两个数据库,分别是Oracle和Mysql,涉及到了多数据源问题,这里做下记录 官方讲解:https:/ ...

  5. tms web core pwa让你的WEB APP离线可用

    tms web core pwa让你的WEB APP离线可用 tms web core允许创建渐进式Web应用程序(PWA).渐进式Web应用程序是为适应在线/离线情况,各种设备类型,最重要的是,让自 ...

  6. 让DELPHI自带的richedit控件显示图片

    让DELPHI自带的richedit控件显示图片 unit RichEx; { 2005-03-04 LiChengbin Added: Insert bitmap or gif into RichE ...

  7. Instrument 实用详解

    苹果:Instruments User Guide iPhone Memory Debugging with NSZombie and Instruments 苹果:Mac OS X Debuggin ...

  8. 【Devops】【Jenkins】Jenkins插件安装失败处理方法

    本篇解释:Jenkins插件安装失败处理方法 不论是刚启动成功后进行的推荐插件安装,还是后期使用Jenkins过程中进行插件的安装.出现插件安装失败的问题,可以通过本篇解决! [注意,插件下载安装失败 ...

  9. Netty 包头

    LengthFieldBasedFrameDecoder 常用的处理大数据分包传输问题的解决类,先对构造方法LengthFieldBasedFrameDecoder中的参数做以下解释说明 maxFra ...

  10. CCBAnimationManager

    #ifndef __CCB_CCBANIMATION_MANAGER_H__ #define __CCB_CCBANIMATION_MANAGER_H__ #include "cocos2d ...