POJ_3009——冰球,IDS迭代加深搜索
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.
- The stone hits a block (Fig. 2(b), (c)).
- 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
#include <cstdio>
int depth;
int w, h, s_x, s_y,
map[][],
dir[][]={,,,-,,,-,}; bool IDS(int d,int x,int y)
{
if(depth == d) //如果搜索深度达到一定,则停止这条搜索树
{
return false;
} for(int i=;i<;i++)
{
int dx = x + dir[i][];
int dy = y + dir[i][]; if(map[dx][dy] == ) //如果下一步直接就撞墙了,中间没有空格,此路就不通
continue; while(!map[dx][dy]) //如果是通路就一直走下去,直到墙壁或者终点
{
dx += dir[i][];
dy += dir[i][];
}
if(dx>= && dx<h && dy>= && dy<w) //如果没有走出地图范围,走出就不用递归了
{
if(map[dx][dy] == ) //撞到的是墙
{
map[dx][dy] = ; //把墙撞碎
if(IDS(d+, dx-dir[i][], dy-dir[i][]))//深度递增,往回走一步
{
return true;
}
map[dx][dy] = ; //墙恢复
}
else //撞到的是终点
{
return true;
}
}
}
return false;
} int main()
{
while(~scanf("%d%d",&w,&h), (w||h))
{
for(int i=;i<h;i++)
{
for(int j=;j<w;j++)
{
scanf("%d",&map[i][j]);
if(map[i][j]==)
{
s_x = i;
s_y = j;
}
}
}
map[s_x][s_y]=; //起点无用,直接覆盖掉
depth = ; //记录搜索深度
while(depth<= && !IDS(, s_x, s_y))
{
depth++;
}
printf(depth==?"-1\n":"%d\n",depth);
}
return ;
}
POJ_3009——冰球,IDS迭代加深搜索的更多相关文章
- POJ1129Channel Allocation[迭代加深搜索 四色定理]
Channel Allocation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14601 Accepted: 74 ...
- BZOJ1085: [SCOI2005]骑士精神 [迭代加深搜索 IDA*]
1085: [SCOI2005]骑士精神 Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 1800 Solved: 984[Submit][Statu ...
- 迭代加深搜索 POJ 1129 Channel Allocation
POJ 1129 Channel Allocation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 14191 Acc ...
- 迭代加深搜索 codevs 2541 幂运算
codevs 2541 幂运算 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamond 题目描述 Description 从m开始,我们只需要6次运算就可以计算出 ...
- HDU 1560 DNA sequence (IDA* 迭代加深 搜索)
题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1560 BFS题解:http://www.cnblogs.com/crazyapple/p/321810 ...
- UVA 529 - Addition Chains,迭代加深搜索+剪枝
Description An addition chain for n is an integer sequence with the following four properties: a0 = ...
- hdu 1560 DNA sequence(迭代加深搜索)
DNA sequence Time Limit : 15000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total ...
- 迭代加深搜索 C++解题报告 :[SCOI2005]骑士精神
题目 此题根据题目可知是迭代加深搜索. 首先应该枚举空格的位置,让空格像一个马一样移动. 但迭代加深搜索之后时间复杂度还是非常的高,根本过不了题. 感觉也想不出什么减枝,于是便要用到了乐观估计函数(O ...
- C++解题报告 : 迭代加深搜索之 ZOJ 1937 Addition Chains
此题不难,主要思路便是IDDFS(迭代加深搜索),关键在于优化. 一个IDDFS的简单介绍,没有了解的同学可以看看: https://www.cnblogs.com/MisakaMKT/article ...
随机推荐
- Java基础知识强化31:String类之String的面试题
1.先看一个图: 2.String面试题: (1)题1: package cn.itcast_02; /* * 看程序写结果 */ public class StringDemo3 { public ...
- 根据goodsId获得相关商品的列表
List<Goods> goodsList = goodsDetailService.getGoodsListByproductId(productId); for (Goods good ...
- 反编译 APKTool 逆向助手
最佳实践--Android逆向助手 1.点击"反编译apk,完成后res下的所有资源就都可以正常使用了,相当于apktool的功能------目前已失效,但是直接用rar解压是可以的!2.点 ...
- C#_DBHelper_SQL数据库操作类.
using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Data; ...
- hdu 2190
//hdu2190 水题 题意是给一个n*3的教室,用1*1,2*2的砖去铺满,有多少种铺法,一开始没发现这个规律,想了一下,应该是递归. #include <iostream> usi ...
- sqlserver中的统计语法
set statisitcs io {on | off} 显示与执行的sql语句有关的磁盘活动量的信息 set statistics profile {on | off} 显示语句的配置文件信息 se ...
- 分析器错误消息: 未能加载类型“WebApplication._Default”
1.新建一个空白解决方案2.新闻一个Web Application项目 默认就有Default.aspx 直接调试的时候出现-------------------------------------- ...
- ajaxfileupload
} } setTimeout( }, s. ...
- 给C++初学者的50个忠告(好文转载)
给C++初学者的50个忠告 1.把C++当成一门新的语言学习(和C没啥关系!真的.): 2.看<Thinking In C++>,不要看<C++变成死相>: 3. ...
- Linux下Fork与Exec使用
Linux下Fork与Exec使用 一.引言 对于没有接触过Unix/Linux操作系统的人来说,fork是最难理解的概念之一:它执行一次却返回两个值.fork函数是Unix系统最杰出的成就之一, ...