POJ 一 3009 Curling 2.0（DFS）

Curling 2.0

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8859 Accepted: 3736

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.

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

题意是扔石头，找到从起点到终点的最短距离，若移动次数超过十次则失败输出-1，若飞出范围就从上一点重新移动。

若是遇到空白可以一直飞，遇到障碍物会把障碍物撞碎并停留在障碍物旁。

使用dfs检测，若遇到障碍物或飞出范围就从上一点换个方向继续移动。

代码如下：

#include<stdio.h>#include<string.h>#define INF 100000000int map[21][21];int dir[4][2]={{0,-1},{1,0},{0,1},{-1,0}};//代表向下、左、上、右移动 int h,w,si,sj,res,step;void dfs(int x,int y);int main(void){    int i,j;    while(scanf("%d%d",&w,&h)==2&&(w+h)){ memset(map,0,sizeof(map)); for(i=0;i<h;i++) {  for(j=0;j<w;j++)  {   scanf("%d",&map[i][j]);   if(map[i][j]==2)   {si=i;//记录下起点和终点     sj=j;   }  } } res=INF; step=0; dfs(si,sj);   if(res<=10) printf("%d\n",res); else printf("-1\n");}return 0;}void dfs(int x,int y){    int a,b,i;    step++;if(step>10||step>=res)return;for(i=0;i<4;i++){ a=x+dir[i][0]; b=y+dir[i][1]; if(a>=h||a<0||b>=w||b<0||map[a][b]==1)      continue;//若离开迷宫或者检测到下一点是障碍物，则换一个方向  while(map[a][b]==2||map[a][b]==0) {  a+=dir[i][0];  b+=dir[i][1];//沿着空白一直走   if(a>=h||a<0||b>=w||b<0)       break;//判断是否离开迷宫  } if(a>=h||a<0||b>=w||b<0)      continue;//若离开迷宫，则换一个方向  if(map[a][b]==3) {  res=step;//到达终点记录下当前步数   return; } if(map[a][b]==1) {  map[a][b]=0;//将障碍物撞碎   dfs(a-dir[i][0],b-dir[i][1]);//若当前点是障碍物，石头停留在上一点处，从上一点开始dfs   step--;//调用一次dfs，step自加1   map[a][b]=1; }}}