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POJ 3009-B - Curling 2.0(DFS)

2016-05-05 13:25 399 查看
B - Curling 2.0
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
3009

Appoint description: 
System Crawler  (2016-05-04)

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 square
1block
2start position
3goal 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


题意:
这题要你找出起点到终点最小的步数,这题石头是会被打碎的,所以用BFS来做会很麻烦,所以还是用DFS做比较好。
AC代码:

/*
这道题一开始就对实例分析错误了,以为是走到边缘就反弹
结果发现写出来一堆bug,最后看了看题解的解析,发现是
自己对题目理解错了。最后在错的基础上重新敲过,最后就
一A了

*/

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstdlib>
using namespace std;
#define T 100005
#define inf 0x3f3f3f3fL
typedef long long ll;

int s[25][25];
int n,m,ma;
int fx[][2]={{1,0},{0,1},{-1,0},{0,-1}};

int jugde(int x,int y,int tx,int ty)
{
int tmp=x;
// x,扫描上下是否存在石块或目的地
if(tx!=0){
while(tmp+tx>=0&&tmp+tx<n&&s[tmp+tx][y]==0){
tmp += tx;
}
//找到目的地
if(tmp+tx>=0&&tmp+tx<n&&s[tmp+tx][y]==3)
return -1;
//找到石块
if(tmp+tx>=0&&tmp+tx<n&&s[tmp+tx][y]!=0)
return tmp;
return 0;
}
// y,扫描左右是否存在石块或目的地
tmp = y;
while(tmp+ty>=0&&tmp+ty<m&&s[x][tmp+ty]==0){
tmp += ty;
}
//找到目的地
if(tmp+ty>=0&&tmp+ty<m&&s[x][tmp+ty]==3)
return -1;
//找到石块
if(tmp+ty>=0&&tmp+ty<m&&s[x][tmp+ty]!=0)
return tmp;
return 0;
}

bool flag;

void dfs(int x,int y,int c)
{
if(c>=10)return;
/*if(s[x][y]==3)ma=min(ma,c);*/
for(int i=0;i<4;++i){
int tx = x+fx[i][0];
int ty = y+fx[i][1];
//位置是否越界
if(tx<0||tx>=n||ty<0||ty>=m||s[tx][ty]==1)continue;
//tmp储存是否存在有石块或目的地的x或y下标
int tmp = jugde(x,y,fx[i][0],fx[i][1]),t;
if(tmp){
if(tmp==-1){
ma = min(ma,c+1);
return;
}
if(fx[i][0]){
if(fx[i][0]>0)t=tmp+1;
else t=tmp-1;
s[t][ty] = 0;
dfs(tmp,ty,c+1);
}
else{
if(fx[i][1]>0)t=tmp+1;
else t=tmp-1;
s[tx][t] = 0;
dfs(tx,tmp,c+1);
}

if(fx[i][0])
s[t][ty] = 1;
else
s[tx][t] = 1;
}
}
}

int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif

int i,j,k,x,y;
while(~scanf("%d%d",&m,&n)&&(n&&m))
{
ma = inf;flag = false;
for(i=0;i<n;++i){
for(j=0;j<m;++j){
scanf("%d",&s[i][j]);
if(s[i][j]==2){
x = i,y = j;
s[i][j] = 0;
}
}
}
dfs(x,y,0);
if(ma==inf)ma=-1;
printf("%d\n",ma);
}
return 0;
}
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