poj 3009 Curling 2.0(DFS)

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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 square mesh(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 willproceed(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 aroute(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

Source

Japan 2006 Domestic

#include<stdio.h>#include<string.h>#include<algorithm>#define inf 0x3f3f3f3fusing namespace std;int a[105][105];int n,m;int x1,y1;int dx[]= {1,0,-1,0};int dy[]= {0,-1,0,1};int mi;int vis[105][105];void dfs(int x,int y,int s)   {    if(a[y][x]==3)    {        mi=min(mi,s);<span style="font-family: 'Times New Roman', Times, serif;">//</span><span style="color:#ff0000;font-family: 'Times New Roman', Times, serif;">DFS只有在全部搜索完后才可以得到最小的步数，每次搜索成功后取当前的最小值</span>        return ;    }    if(s>=10)//       <span style="color:#ff6666;">步数大于大于10，题目中说按没有搜索到，等于10，如果满足的话会在上面返回。</span>    {        return ;    }    if(x+1<n&&(a[y][x+1]!=1||vis[y][x+1]))          <span style="color:#ff0000;">//判断在此行是否可以移动，可移动的条件是在这一行找到0（障碍物）或3（目标）否则一</span>

<span style="color:#ff0000;">                                                                             //  旦移动将不会停下来或者是yuejie</span>    {        int i;        if(a[y][x+1]==3)                                                 {           dfs(x+1,y,s+1);<span style="font-family: 'Times New Roman', Times, serif;">                  </span><span style="color:#ff0000;font-family: 'Times New Roman', Times, serif;">//如果当前的下一个位置是目标则进入dfs</span>        }        if(vis[y][x+1])                         {                                                  i=x+1;        }        else i=x+2;        for(; i<n; i++)        {            if((a[y][i]==1||a[y][i]==3)&&!vis[y][i])            {                if(a[y][i]==1)                {                    vis[y][i]=1;                    dfs(i-1,y,s+1);                    vis[y][i]=0;                }                else                {                    vis[y][i]=1;                    dfs(i,y,s+1);                    vis[y][i]=0;                }                break;            }        }    }    if(x-1>=0&&(a[y][x-1]!=1||vis[y][x-1]))    {        int i;        if(a[y][x-1]==3)        {            dfs(x-1,y,s+1);        }        if(vis[y][x-1])i=x-1;        else i=x-2;        for(; i>=0; i--)        {            if((a[y][i]==1||a[y][i]==3)&&!vis[y][i])            {                if(a[y][i]==1)                {                    vis[y][i]=1;                    dfs(i+1,y,s+1);                    vis[y][i]=0;                }                else                {                    vis[y][i]=1;                    dfs(i,y,s+1);                    vis[y][i]=0;                }                break;            }        }    }    if(y-1>=0&&(a[y-1][x]!=1||vis[y-1][x]))    {        int j;        if(a[y-1][x]==3)        {            dfs(x,y-1,s+1);        }        if(vis[y-1][x])j=y-1;        else j=y-2;        for(; j>=0; j--)        {            if((a[j][x]==1||a[j][x]==3)&&!vis[j][x])            {                if(a[j][x]==1)                {                    vis[j][x]=1;                    dfs(x,j+1,s+1);                    vis[j][x]=0;                }                else                {                    vis[j][x]=1;                    dfs(x,j,s+1);                    vis[j][x]=0;                }                break;            }        }    }    if(y+1<m&&(a[y+1][x]!=1||vis[y+1][x]))    {         int j;         if(a[y+1][x]==3)         {dfs(x,y+1,s+1);         }         if(vis[y+1][x])j=y+1;         else j=y+2;        for(; j<m; j++)        {            if((a[j][x]==1||a[j][x]==3)&&!vis[j][x])            {                if(a[j][x]==1)                {                    vis[j][x]=1;                    dfs(x,j-1,s+1);                    vis[j][x]=0;                }                else                {                    vis[j][x]=1;                    dfs(x,j,s+1);                    vis[j][x]=0;                }                break;            }        }    }}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)break;        mi=inf;        for(int i=0; i<m; i++)            for(int j=0; j<n; j++)            {                scanf("%d",&a[i][j]);                if(a[i][j]==2)                {                    x1=j;                    y1=i;                }            }        int t=0;        int tt=0;       // printf("%d %d\n",x1,y1);        for(int i=0; i<4; i++)        {            int xi=x1+dx[i];            int yi=y1+dy[i];            if(xi>=0&&yi>=0&&xi<n&&yi<m)            {                if(a[yi][xi]==0)                {                    t=1;                }                if(a[yi][xi]==3)                {                    tt=1;                }            }        }        if(tt)printf("1\n");        else        if(!t)printf("-1\n");        else        {            memset(vis,0,sizeof(vis));            dfs(x1,y1,0);            if(mi==inf)printf("-1\n");            else printf("%d\n",mi);        }    }}

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