poj3009 Curling 2.0

Description

On Planet MM-21, after their Olympic, the year, and curling is getting popular. But the rules is somewhat different from ours. The game is played on the Ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to leads 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 is occupied with blocks. There is special squares namely the start and the goal, which is not occupied with blocks. (These squares is distinct.) Once The stone begins to move, it'll proceed until it hits a block. In order to bring the stone to the goal, your may has 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 is restricted to X and y directions. Diagonal moves is prohibited.
• When the stone stands still, you can make it moving by throwing it. 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 hits.
    • 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 to more than times in a game. If The stone does not reach the goal in 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 O F moves required.

With the initial configuration shown in Fig. 1, 4 moves is 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 had 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 to a line containing and a 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 <=, 1 <= h <= 20.

Each line consists of the 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 has a decimal integer indicating the minimum number of moves along a route from the star T to the goal. If There is no such routes, print-1 instead. Each of the line should is not a 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
This problem can be done with deep search. Wide search to think it is very difficult to achieve. Mainly to pay attention to a direction to the end of the realization, here need a lot of inference, the other is the basis of the deep search writing, attention to pruning.
#include <iostream> #include <stdio.h> #include <string.h> #include <math.h> #include < vector> #include <queue> #include <stack> #include <string> #include <algorithm>using namespace Std;int map[30][30],t;int tab[10][2]={0,0,0,1,-1,0,0,-1,1,0};int x2,y2,x3,y3,n,m;void dfs (int x,int y,int Time) {int i,j,xx,yy,flag1;if (time>10) return;if (time==10 && (x!=x3 && y!=y3)) return;if (X==x3 & & Y==y3) {if (t>time) T=time;return;} if (time>=t) return;for (i=1;i<=4;i++) {flag1=1;xx=x;yy=y;if (map[x+tab[i][0]][y+tab[i][1]]==1) continue;while (1) {xx=xx+tab[i][0];yy=yy+tab[i][1];if (xx==x3 && yy==y3) break;if (xx<1 | | xx>n | | yy<1 | | yy>m) {flag1= 0;break;} if (map[xx][yy]==0) continue;if (map[xx][yy]==1) {flag1=2;xx=xx-tab[i][0];yy=yy-tab[i][1];break;}} if (flag1==0) continue;//printf ("%d%d\n", xx,yy), if (flag1==2) {Map[xx+tab[i][0]][yy+tab[i][1]]=0;dfs (xx,yy,time+1) ; map[xx+tab[i][0]][yy+tab[i][1]]=1;} Else{dfs (xx,yy,time+1);}}} int main ({int I,j;while (scanf ("%d%d", &m,&n)!=eof) {if (n==0 && m==0) break;memset (map,0,sizeof (map)); for (I=1 ; i<=n;i++) {for (j=1;j<=m;j++) {scanf ("%d", &map[i][j]), if (map[i][j]==2) {x2=i;y2=j;} else if (map[i][j]==3) {x3=i;y3=j;}}} T=12;dfs (x2,y2,0); if (t==12) {printf (" -1\n");} else printf ("%d\n", t);} return 0;}

poj3009 Curling 2.0