Poj 3009 Curling 2.0【dfs】
来源:互联网 发布:udp端口命令 编辑:程序博客网 时间:2024/06/11 12:48
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.
0vacant square1block2start position3goal 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 13 26 61 0 0 2 1 01 1 0 0 0 00 0 0 0 0 30 0 0 0 0 01 0 0 0 0 10 1 1 1 1 16 11 1 2 1 1 36 11 0 2 1 1 312 12 0 1 1 1 1 1 1 1 1 1 313 12 0 1 1 1 1 1 1 1 1 1 1 30 0
Sample Output
14-1410-1
题意说的是,给出一个起点和一个终点,物体只有碰到冰块或者到达终点的时候才会停下,而且被碰到的冰块会消失,现在给出一个图,问最少推动多少下,才能到达终点。
Dfs 回溯+剪枝.....
特别需要注意的一点是,直接和物体相邻的冰块不能进行撞击
#include<stdio.h>#include<string.h>const int maxn=55;#define min(a,b) (a>b?b:a)int n,m,map[55][55];int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1},ans;int out(int a,int b)//出界判断 {return a<0||a>=n||b<0||b>=m;}void dfs(int x,int y,int cnt){if(cnt>10){return;}for(int i=0;i<4;++i)//尝试四个方向的前进 {int tx=x+dx[i],ty=y+dy[i]; if(out(tx,ty)||map[tx][ty]==1)//出界或者有挡住的 {continue;}while(!out(tx,ty)&&map[tx][ty]==0)//否则只要不出界,一直前进到遇到冰块 {tx+=dx[i];ty+=dy[i];}if(out(tx,ty))//是因为出界停止的 {continue;}else if(map[tx][ty]==3){ans=min(ans,cnt);return;}map[tx][ty]=0;dfs(tx-dx[i],ty-dy[i],cnt+1);map[tx][ty]=1; }}int main(){//freopen("shuju.txt","r",stdin);while(scanf("%d%d",&m,&n),m|n){int bx,by;for(int i=0;i<n;++i){for(int j=0;j<m;++j){scanf("%d",&map[i][j]);if(map[i][j]==2){bx=i;by=j;map[i][j]=0;//把起点当成一般的点}}}ans=maxn;dfs(bx,by,1);printf("%d\n",ans>10?-1:ans);}return 0;}
- POJ 3009 Curling 2.0 DFS
- poj-3009-Curling 2.0-dfs
- poj 3009 Curling 2.0 DFS
- poj 3009 Curling 2.0 DFS
- POJ 3009Curling 2.0(DFS)
- POJ 3009 Curling 2.0 (DFS)
- poj-3009-Curling 2.0-dfs
- poj 3009 Curling 2.0 DFS
- POJ 3009 Curling 2.0 (DFS)
- poj 3009 Curling 2.0 【DFS】
- POJ 3009 Curling 2.0 DFS
- poj 3009 Curling 2.0 (dfs)
- Curling 2.0 (poj 3009 dfs)
- POJ 3009--Curling 2.0【DFS】
- POJ 3009 Curling 2.0(DFS)
- poj-3009 Curling 2.0-DFS
- POJ-3009-Curling 2.0- DFS
- POJ 3009 Curling 2.0 (DFS)
- 第4章 Android dex文件格式 第二节
- 关于Android Studio里的Gradle,你所需要知道的都在这里了
- 美团Android资源混淆保护实践
- C# 枚举与结构
- Java基础之集合
- Poj 3009 Curling 2.0【dfs】
- ubuntu 14.04 开启root登录
- Spark代码2之Transformation:union,distinct,join
- Android学习笔记----利用ContentObserver监听数据库内容的改变
- Oracle 11gR2 SCAN 详解
- Android 布局中 如何使控件居中
- is_recovery_dest_file
- ScrollView分析
- 使用Gradle构建Android项目的一些自定义配置