POJ-3009-Curling 2.0

Curling 2.0

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17782 Accepted: 7315

Description

On Planet MM-21, after their Olympic games this year, curling(卷曲) is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a squaremesh(网眼) 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(最小的) number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied(占据) with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct(明显的).) Once the stone begins to move, it will proceed(开始) 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(斜的) moves are prohibited(阻止).
• 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(最小的) number of moves required.

With the initial configuration(配置) shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route(路线) 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(解决方案) for Fig. D-1 and the final board configuration

Input

The input(投入) is a sequence(序列) of datasets. The end of the input is indicated(表明) by a line containing two zeros separated by a space. The number of datasets never exceeds(超过) 100.

Each dataset is formatted(格式化) 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(小数) numbers delimited(划界) 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(小数的) integer(整数) indicating(表明) the minimum(最小的) number of moves along a route(路线) 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

给出地图,0代表空地,1是障碍物, 2是起始点,3是目标点,要求将冰球打到目标点, 冰球紧贴墙的话球到墙的方向不能打, 球碰不到墙会跑到地图外,撞到墙会停在墙前方,被撞的那一格墙回消失,求冰球到目标点的操作次数,无法到达输出-1

DFS:

#include<iostream>#include<string.h>#include<stdio.h>#include<math.h>#include<algorithm>#include<queue>#include<vector>using namespace std;int w, h, step;struct node{    int x, y;    int Map[21][21];} p, pp;bool kk(int x, int y){    if(x>=0&&y>=0&&x<h&&y<w)return true;     return false;}void dfs(node p, int s){    if(s>=step)return;    int x, y;    // up    x = p.x-1, y = p.y;    if(kk(x,y)&&p.Map[x][y]!=1)    {        while(kk(x,y)&&p.Map[x][y]==0)--x;        if(kk(x,y))        {            if(p.Map[x][y]==3)            {                if(step>s+1)step = s+1;                return;            }            pp = p;            pp.x = x+1;            pp.y = y;            pp.Map[x][y] = 0;            dfs(pp, s+1);        }    }   // down    x = p.x+1, y = p.y;    if(kk(x,y)&&p.Map[x][y]!=1)    {        while(kk(x,y)&&p.Map[x][y]==0)++x;        if(kk(x,y))        {            if(p.Map[x][y]==3)            {                if(step>s+1)step = s+1;                return;            }            pp = p;            pp.x = x-1;            pp.y = y;            pp.Map[x][y] = 0;            dfs(pp, s+1);        }    }    // left    x = p.x, y = p.y-1;    if(kk(x,y)&&p.Map[x][y]!=1)    {        while(kk(x,y)&&p.Map[x][y]==0)--y;        if(kk(x,y))        {            if(p.Map[x][y]==3)            {                if(step>s+1)step = s+1;                return;            }            pp = p;            pp.x = x;            pp.y = y+1;            pp.Map[x][y] = 0;            dfs(pp, s+1);        }    }     // right    x = p.x, y = p.y+1;    if(kk(x,y)&&p.Map[x][y]!=1)    {        while(kk(x,y)&&p.Map[x][y]==0)++y;        if(kk(x,y))        {            if(p.Map[x][y]==3)            {                if(step>s+1)step = s+1;                return;            }            pp = p;            pp.x = x;            pp.y = y-1;            pp.Map[x][y] = 0;            dfs(pp, s+1);        }    }}int main(){    while(~scanf("%d%d", &w, &h)&&(w||h))    {        for(int i = 0; i<h; ++i)        {            for(int j = 0; j<w; ++j)            {                scanf("%d",&p.Map[i][j]);                if(p.Map[i][j]==2)                {                    p.x = i;                    p.y = j;                    p.Map[i][j] = 0;                }            }        }        step = 11;        dfs(p, 0);        if(step==11)step = -1;        printf("%d\n", step);    }    return 0;}