SDAU课程练习2 1022

Pusher

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/65536K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 2

Special Judge

Problem Description

PusherBoy is an online game http://www.hacker.org/push . There is an R * C grid, and there are piles of blocks on some positions. The goal is to clear the blocks by pushing into them. <br><br>You should choose an empty area as the initial position of the PusherBoy.
Then you can choose which direction (U for up, D for down, L for left and R for right) to push. Once the direction is chosen, the PusherBoy will walk ahead until he met a pile of blocks (Walking outside the grid is invalid). Then he remove one block from the
pile (so if the pile contains only one block, it will become empty), and push the remaining pile of blocks to the next area. (If there have been some blocks in the next area, the two piles will form a new big pile.)<br><br>Please note if the pusher is right
up against the block, he can't remove and push it. That is, there must be a gap between the pusher and the pile. As the following figure, the pusher can go up, but cannot go down. (The cycle indicates the pusher, and the squares indicate the blocks. The nested
squares indicate a pile of two blocks.)<br><center><img src=../../../data/images/2821-1.jpg></center> <br><br>And if a whole pile is pushed outside the grid, it will be considered as cleared.<br>

 

Input

There are several test cases in each input. The first two lines of each case contain two numbers C and R. (R,C <= 25) Then R lines follow, indicating the grid. '.' stands for an empty area, and a lowercase letter stands for a pile of blocks. ('a' for one block,
'b' for two blocks, 'c' for three, and so on.)<br><br>

 

Output

Output three lines for each case. The first two lines contains two numbers x and y, indicating the initial position of the PusherBoy. (0 <= x < R, 0 <= y < C). The third line contains a moving sequence contains 'U', 'D', 'L' and 'R'. Any correct answer will
be accepted.

 

Sample Input

3<br>7<br>...<br>...<br>.b.<br>...<br>...<br>.a.<br>...<br>

 

Sample Output

4<br>1 UDU

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

题目大意：

上面有链接，可以去玩一下这个游戏再来看题。。

可以上下左右移动，移动遇到箱子终止，然后把箱子朝着当前方向推出一格，箱子数量减少  1  

限制条件，当前位置要和箱子有一个距离才推得动箱子。

要求输出的是起点坐标和操作。

思路：

感觉用到的模拟更多一些，  dfs  进行了起点坐标的检索和推动方向的检索，大概是这样吧。

感想：

参考别人博客才写出来的，  dfs  感觉遇到了一个卡住的地方。
还有就是  dfs  不只是 回溯，一开始想着回溯来着，但是写不出来，也可能这个题不是“严格的dfs”
我觉得还是和贪心一样，  dfs  也只是一个思想，结合其他的在一起用才能发挥出来效果，

AC代码：

#include<cstdio>
#include<cstring>
#include<cctype>

int n,m,num,cnt,a[35][35],dirx[4]={0,0,-1,1},diry[4]={-1,1,0,0};
char mapp[35][35],p[]="LRUD",path[625];
int isend(int x,int y)
{
if(x<0||y<0||x>=n||y>=m)
return 0;
if(a[x][y]==0)
return 1;
return -1;
}
int dfs(int x,int y,int pos)
{
int i,xx,yy,tx,ty;
for(i=0;i<4;i++)
{
xx=x+dirx[i];
yy=y+diry[i];
if(isend(xx,yy)!=1)
continue;
do
{
xx+=dirx[i];
yy+=diry[i];
}while(isend(xx,yy)==1);
if(isend(xx,yy)&&isend(tx=xx+dirx[i],ty=yy+diry[i])&&a[xx][yy])
{
int t=a[xx][yy];
path[pos]=p[i];
if(t==1||a[tx][ty])
cnt--;
if(cnt==0)
{
num=pos;
return 1;
}
a[tx][ty]+=t-1;
a[xx][yy]=0;
if(dfs(xx,yy,pos+1))
return 1;
a[xx][yy]=t;
a[tx][ty]-=t-1;
if(t==1||a[tx][ty])
cnt++;
}
}
return 0;
}
int ma
4000
in()
{
// freopen("r.txt","r",stdin);
while(~scanf("%d%d",&m,&n))
{
int i,j,k,x,y,flag=1,f[35][35]={0};
for(i=num=0;i<n;i++)
{
scanf("%s",mapp[i]);
for(j=0;j<m;j++)
if(isalpha(mapp[i][j]))
num++,a[i][j]=mapp[i][j]-'a'+1;
else
a[i][j]=0;
}
for(i=0;i<n&&flag;i++)
for(j=0;j<m&&flag;j++)
if(a[i][j])
for(k=0;k<4&&flag;k++)
{
x=i+dirx[k]*2;
y=j+diry[k]*2;
if(isend(x,y)>0&&!f[x][y])
{
f[x][y]=1;
cnt=num;
if(dfs(x,y,0))
flag=0;
}
}
printf("%d\n%d\n",x,y);
for(i=0;i<=num;i++)
printf("%c",path[i]);
puts("");
}
}