poj 3009 Curling 2.0

Curling 2.0

      

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大致题意：2为起点，3为终点，0为空地，1为石头，我们要从起点搜到终点；需要注意的是：要看该冰壶是运动的还是静止的，如果冰壶是静止的，恰巧它附近又是石头，那么它就被堵在原地，无法到达终点，如果它是运动的，如果它前方有石头，它就会把前面的那个石头撞碎（即该点由1变为0），同时它也会停在石头的前面，若前面全为空地，则它会沿着路经一直走下去，直到出界或者撞上石头（该题申明：当移动的步数超过10时，我们自动认为这次游戏失败了）附上代码和详细解释：#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <queue>#include <iomanip>using namespace std;struct node{    int x,y;    int map[22][22];};int m,n;int step;void dfs(node a,int k){    if(k>=step) return;//如果步数大于题目所给的要求，直接返回    int i;    if(a.x+1<n && a.map[a.x+1][a.y]!=1)//向下找，且无出界，下面也不是石头    {        for(i=1;a.x+i<n;i++)//此时石头是在移动的        {            if(a.map[a.x+i][a.y]==1)//如果前方有石头，则让石头变为空地            {                node b=a;                b.map[a.x+i][a.y]=0;//石头变为空地                b.x=a.x+i-1;//此时冰壶停在石头的正上方                dfs(b,k+1);                break;            }            if(a.map[a.x+i][a.y]==3)//如果到了终点            {                if(step>k+1)//寻找最小步数                {                    step=k+1;                }                return;            }        }    }    if(a.x-1>=0 && a.map[a.x-1][a.y]!=1)//向上找，思路和上面的是一样的    {        for(i=1;a.x-i>=0;i++)        {            if(a.map[a.x-i][a.y]==1)            {                node b=a;                b.map[a.x-i][a.y]=0;                b.x=a.x-i+1;//此时冰壶是在石头的正下方                dfs(b,k+1);                break;            }            if(a.map[a.x-i][a.y]==3)            {                if(step>k+1)                {                    step=k+1;                }                return;            }        }    }    if(a.y-1>=0 && a.map[a.x][a.y-1]!=1)//向左找    {        for(i=1;a.y-i>=0;i++)        {            if(a.map[a.x][a.y-i]==1)            {                node b=a;                b.map[a.x][a.y-i]=0;                b.y=a.y-i+1;//此时冰壶是在石头的右方                dfs(b,k+1);                break;            }            if(a.map[a.x][a.y-i]==3)            {                if(step>k+1)                {                    step=k+1;                }                return;            }        }    }    if(a.y+1<m && a.map[a.x][a.y+1]!=1)//向右找    {        for(i=1;a.y+i<m;i++)        {            if(a.map[a.x][a.y+i]==1)            {                node b=a;                b.map[a.x][a.y+i]=0;                b.y=a.y+i-1;//此时冰壶是在石头的左方                dfs(b,k+1);                break;            }            if(a.map[a.x][a.y+i]==3)            {                if(step>k+1)                {                    step=k+1;                }                return;            }        }    }}int main(){     int i,j;     while(cin>>m>>n)     {         if(m==0) break;         node p;         memset(p.map,0,sizeof(p.map));         for(i=0;i<n;i++)         {             for(j=0;j<m;j++)             {                 cin>>p.map[i][j];                 if(p.map[i][j]==2)                 {                     p.x=i;                     p.y=j;                 }             }         }         step=11;//因为题目说明，大于10步，则算游戏失败，那么不是大于等于11，就是失败         dfs(p,0);         if(step==11)         {             cout<<"-1"<<endl;         }else{             cout<<step<<endl;         }     }     return 0;      }