poj 3009 Curling 2.0 【dfs经典题目】

Curling 2.0

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13611		Accepted: 5692

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.

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 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 1
3 2
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
6 1
1 1 2 1 1 3
6 1
1 0 2 1 1 3
12 1
2 0 1 1 1 1 1 1 1 1 1 3
13 1
2 0 1 1 1 1 1 1 1 1 1 1 3
0 0

Sample Output

1
4
-1
4
10
-1

题目意思：把一个冰壶从 2 位置用最少的步数移动到 3 位置。 1代表障碍物，0代表空白区域。

遵循规则：冰壶一旦沿着一个方向前进就不能停下，除非碰撞到障碍物才能停下，碰撞的同时障碍物消失，冰壶停在障碍前一位置，继续下一次方向选择。

注意：1,冰壶每次选择的方向是没有障碍的方向; 2,冰壶越界也算输。

写程序不到1小时，调试2个多小时。。。 错了很多次，醉了。

先看个错误的程序：

#include <cstdio>
#include <cstring>
#include <algorithm> 
#define INF 0x3f3f3f3f
#define MAX 22
using namespace std;
int map[MAX][MAX];
int n, m;
int sx, sy, ex, ey;
int ans;//最小步数 
int move[4][2] = {0,1, 0,-1, 1,0, -1,0};
int judge(int x, int y)//判断是否越界 
{
    if(x >= 0 && x < n && y >= 0 && y < m)
    return 1;
    else
    return 0;
}
void getmap()
{
    memset(map, 0, sizeof(map));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            scanf("%d", &map[i][j]);
            if(map[i][j] == 2)
            {
                sx = i;
                sy = j;
                map[i][j] = 0;
            }
            if(map[i][j] == 3)
            {
                ex = i;
                ey = j;
                map[i][j] = 0;
            }
        }
    }
}
void dfs(int x, int y, int step)
{
    int k;
    int next_x, next_y;//下一次搜索坐标  
    if(step >= 10) return ;
    for(k = 0; k < 4; k++)
    {
        next_x = x + move[k][0];
        next_y = y + move[k][1];
        if(map[next_x][next_y] || !judge(next_x, next_y))//有障碍  或者 越界 
        continue;
        while(!map[next_x][next_y])//一直前进 直到碰到障碍 
        {
            if(next_x == ex && next_y == ey)
            {
                ans = min(ans, step+1);
                return ;
            }
            if(!judge(next_x, next_y))//越界  
            break;
            next_x += move[k][0];//继续前进 
            next_y += move[k][1];
        }
        /*越界的话当前位置一定为0,没越界的情况下才可能是1且上一坐标不会越界 所以不用判断上一坐标是否越界*/
        if(map[next_x][next_y])//碰到障碍
        {
            map[next_x][next_y] = 0;
            dfs(next_x-move[k][0], next_y-move[k][1], step+1);
            map[next_x][next_y] = 1;
        } 
    } 
}
int main()
{
    while(scanf("%d%d", &m, &n), m||n)
    {
        getmap();
        ans = INF;
        dfs(sx, sy, 0);
        if(ans == INF)
        printf("-1\n");
        else
        printf("%d\n", ans);
    }
    return 0;
}

代码错误的原因是这一部分：

/*越界的话当前位置一定为0,没越界的情况下才可能是1且上一坐标不会越界 所以不用判断上一坐标是否越界*/
        if(map[next_x][next_y])//碰到障碍
        {
            map[next_x][next_y] = 0;
            dfs(next_x-move[k][0], next_y-move[k][1], step+1);
            map[next_x][next_y] = 1;
        }

我很费解为什么，当前位置为1的坐标一定不会越界的，为什么加上个judge(next_x, next_y)才过。。。求指教



这个是ac的代码：110ms

#include <cstdio>
#include <cstring>
#include <algorithm> 
#define INF 0x3f3f3f3f
#define MAX 22
using namespace std;
int map[MAX][MAX];
int n, m;
int sx, sy, ex, ey;
int ans;//最小步数 
int move[4][2] = {0,1, 0,-1, 1,0, -1,0};
int judge(int x, int y)//判断是否越界 
{
    if(x >= 0 && x < n && y >= 0 && y < m)
    return 1;
    else
    return 0;
}
void getmap()
{
    memset(map, 0, sizeof(map));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            scanf("%d", &map[i][j]);
            if(map[i][j] == 2)
            {
                sx = i;
                sy = j;
                map[i][j] = 0;
            }
            if(map[i][j] == 3)
            {
                ex = i;
                ey = j;
                map[i][j] = 0;
            }
        }
    }
}
void dfs(int x, int y, int step)
{
    int k;
    int next_x, next_y;//下一次搜索坐标  
    if(step >= 10) return ;
    for(k = 0; k < 4; k++)
    {
        next_x = x + move[k][0];
        next_y = y + move[k][1];
        if(map[next_x][next_y] || !judge(next_x, next_y))//有障碍  或者 越界 
        continue;
        while(!map[next_x][next_y])//一直前进 直到碰到障碍 
        {
            if(next_x == ex && next_y == ey)
            {
                ans = min(ans, step+1);
                return ;
            }
            if(!judge(next_x, next_y))//越界  
            break;
            next_x += move[k][0];//继续前进 
            next_y += move[k][1];
        }
        /*越界的话当前位置一定为0,没越界的情况下才可能是1且上一坐标不会越界 所以不用判断上一坐标是否越界*/
        if(map[next_x][next_y] && judge(next_x, next_y))//碰到障碍
        {
            map[next_x][next_y] = 0;
            dfs(next_x-move[k][0], next_y-move[k][1], step+1);
            map[next_x][next_y] = 1;
        } 
    } 
}
int main()
{
    while(scanf("%d%d", &m, &n), m||n)
    {
        getmap();
        ans = INF;
        dfs(sx, sy, 0);
        if(ans == INF)
        printf("-1\n");
        else
        printf("%d\n", ans);
    }
    return 0;
}

另一个：157ms：

#include <cstdio>
#include <cstring>
#include <algorithm> 
#define INF 0x3f3f3f3f 
#define MAX 22
using namespace std;
int map[MAX][MAX];
int n, m;
int sx, sy, ex, ey;
int ans;//最小步数 
int move[4][2] = {0,1, 0,-1, 1,0, -1,0};
int judge(int x, int y)//判断是否越界 
{
    if(x >= 0 && x < n && y >= 0 && y < m)
    return 1;
    else
    return 0;
}
void getmap()
{
    memset(map, 0, sizeof(map));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            scanf("%d", &map[i][j]);
            if(map[i][j] == 2)
            {
                sx = i;
                sy = j;
                map[i][j] = 0;
            }
            if(map[i][j] == 3)
            {
                ex = i;
                ey = j;
                map[i][j] = 0;
            }
        }
    }
}
void dfs(int x, int y, int step)
{
    int k;
    int next_x, next_y;//下一次搜索坐标  
    if(step >= 10) return ;//剪枝  
    for(k = 0; k < 4; k++)
    {
        next_x = x + move[k][0];
        next_y = y + move[k][1];
        if(map[next_x][next_y] || !judge(next_x, next_y))//有障碍  或者 越界 
        continue;
        next_x = x;
        next_y = y;
        while(!map[next_x][next_y] && judge(next_x, next_y))//一直前进 直到碰到障碍 或者越界 
        {
            next_x += move[k][0];//继续前进 
            next_y += move[k][1];
            if(next_x == ex && next_y == ey)
            {
                ans = min(ans, step+1);
                return ;
            }
        }
        /*越界的话当前位置一定为0,没越界的情况下才可能是1且上一坐标不会越界 所以不用判断上一坐标是否越界*/
        if(map[next_x][next_y] && judge(next_x, next_y))//碰到障碍 且不能越界 
        {
            map[next_x][next_y] = 0;
            dfs(next_x-move[k][0], next_y-move[k][1], step+1);
            map[next_x][next_y] = 1;
        } 
    } 
}
int main()
{
    while(scanf("%d%d", &m, &n), m||n)
    {
        getmap();
        ans = INF;
        dfs(sx, sy, 0);
        if(ans == INF)
        printf("-1\n");
        else
        printf("%d\n", ans);
    }
    return 0;
}