eoj1857

push!

Time Limit:1000MSMemory Limit:30000KB
Total Submit:124Accepted:67

Description

Mr. Schwarz was a famous powerful pro wrestler. He starts a part time job as a warehouseman. His task is to move a cargo to a goal by repeatedly pushing the cargo in the warehouse, of course, without breaking the walls and the pillars of the warehouse.

There may be some pillars in the warehouse. Except for the locations of the pillars, the floor of the warehouse is paved with square tiles whose size fits with the cargo. Each pillar occupies the same area as a tile.


Initially, the cargo is on the center of a tile. With one push, he can move the cargo onto the center of an adjacent tile if he is in proper position. The tile onto which he will move the cargo must be one of (at most) four tiles (i.e., east, west, north or south) adjacent to the tile where the cargo is present.

To push, he must also be on the tile adjacent to the present tile. He can only push the cargo in the same direction as he faces to it and he cannot pull it. So, when the cargo is on the tile next to a wall (or a pillar), he can only move it along the wall (or the pillar). Furthermore, once he places it on a corner tile, he cannot move it anymore.

He can change his position, if there is a path to the position without obstacles (such as the cargo and pillars) in the way. The goal is not an obstacle. In addition, he can move only in the four directions (i.e., east, west, north or south) and change his direction only at the center of a tile.

As he is not so young, he wants to save his energy by keeping the number of required pushes as small as possible. But he does not mind the count of his pedometer, because walking is very light exercise for him.

Your job is to write a program that outputs the minimum number of pushes required to move the cargo to the goal, if ever possible.

Input

The input consists of multiple maps, each representing the size and the arrangement of the warehouse. A map is given in the following format.

w h
d11 d12 d13 ... d1w
d21 d22 d23 ... d2w
. . .
dh1 dh2 dh3 ... dhw

The integers w and h are the lengths of the two sides of the floor of the warehouse in terms of widths of floor tiles. w and h are less than or equal to 7. The integer dij represents what is initially on the corresponding floor area in the following way.

0: nothing (simply a .oor tile)
1: a pillar
2: the cargo
3: the goal
4: the warehouseman (Mr. Schwarz)

Each of the integers 2, 3 and 4 appears exactly once as dij in the map. Integer numbers in an input line are separated by at least one space character. The end of the input is indicated by a line containing two zeros.

Output

For each map, your program should output a line containing the minimum number of pushes. If the cargo cannot be moved to the goal, -1 should be output instead.

Sample Input

5 5
0 0 0 0 0
4 2 0 1 1
0 1 0 0 0
1 0 0 0 3
1 0 0 0 0
5 3
4 0 0 0 0
2 0 0 0 0
0 0 0 0 3
7 5
1 1 4 1 0 0 0
1 1 2 1 0 0 0
3 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 1 0 0 0
6 6
0 0 0 0 0 3
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 2 0 0 0 0
4 0 0 0 0 0
0 0

Sample Output

5
-1
11

8


题目:EOJ1857

 

题目分析:用两层bfs,主bfs表示箱子到达目的点的最短路径,若用dx[]={-1,1,0,0},dy[]={0,0,-1,1}表示移动方向,now表示当前节点,next表示下一节点(next.x=now.x+dx[i];next.y=now.y+dy[i];),若能从now访问到next节点需满足条件:搬运工能移动到点(now.x-dx[i],now-dy[i]),则可用bfs_man另一个bfs函数判断此条件能否成立。同样,应注意到对箱子来说每个点可从四个方向访问。

 

AC代码:

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <queue>

 

using namespace std;

int map[10][10];

boolvisited[10][10][4],visited_man[10][10];

int dx[]={-1,1,0,0},dy[]={0,0,-1,1};

 

struct Node{

   int x,y,step,x_man,y_man;

   Node(){x=y=step=x_man=y_man=0;}

};

 

bool bfs_man(int x,int y,int x_goal,inty_goal,int w,int h,int x_full,int y_full){

   Node now,next;

   now.x=x;now.y=y;

   memset(visited_man,false,sizeof(visited_man));

   visited_man[x][y]=true;

   queue<Node> q;

   q.push(now);

   while(!q.empty()){

       now=q.front();

       q.pop();

       if(now.x==x_goal && now.y==y_goal)

           return true;

       for(int i=0;i<4;++i){

           next.x=now.x+dx[i];next.y=now.y+dy[i];

           if(next.x>=0 && next.x<h && next.y>=0&& next.y<w && !visited_man[next.x][next.y] &&map[next.x][next.y]!=1 && (next.x!=x_full || next.y!=y_full)){

                if(next.x==x_goal &&next.y==y_goal)

                    return true;

                q.push(next);

               visited_man[next.x][next.y]=true;

           }

       }

    }

   return false;

}

 

int bfs(int x,int y,int x_goal,inty_goal,int x_man,int y_man,int w,int h){

   Node now,next;

   now.x=x;now.y=y;

   now.x_man=x_man;now.y_man=y_man;

   queue<Node> q;

   q.push(now);

   while(!q.empty()){

       now=q.front();

       q.pop();

       for(int i=0;i<4;++i){

           next.x=now.x+dx[i];

           next.y=now.y+dy[i];

           next.x_man=now.x;

           next.y_man=now.y;

           next.step=now.step+1;

           int to_x=now.x-dx[i],to_y=now.y-dy[i];

 

           memset(visited_man,false,sizeof(visited_man));

           bool tem=bfs_man(now.x_man,now.y_man,to_x,to_y,w,h,now.x,now.y);

           if(next.x>=0 && next.x<h && next.y>=0&& next.y<w && !visited[next.x][next.y][i] &&map[next.x][next.y]!=1

               && tem ){

                if(next.x==x_goal &&next.y==y_goal){

                    return next.step;

                }

                q.push(next);

                visited[next.x][next.y][i]=true;

           }

       }

    }

   return -1;

}

 

int main()

{

   int w,h,i,j,x,y,x_man,y_man,x_goal,y_goal;

   while(scanf("%d%d",&w,&h),h||w){

       memset(visited,false,sizeof(visited));

       memset(visited_man,false,sizeof(visited_man));

       for(i=0;i<h;++i){

           for(j=0;j<w;++j){

               scanf("%d",&map[i][j]);

                if(map[i][j]==2){

                    x=i;y=j;

                }

                if(map[i][j]==4){

                    x_man=i;y_man=j;

                }

                if(map[i][j]==3){

                    x_goal=i;y_goal=j;

                }

           }

       }

//       cout<<bfs_man(x_man,y_man,x_goal,y_goal,w,h,x,y)<<endl;

       int ans=bfs(x,y,x_goal,y_goal,x_man,y_man,w,h);

       printf("%d\n",ans);

    }

   return 0;

}


### 关于EOJ DNA排序问题的解题思路 在处理EOJ中的DNA排序问题时,主要挑战在于如何高效地完成字符串数组的排序以及去重操作。由于题目涉及两个测试点可能因时间复杂度较高而超时,因此需要优化算法设计。 #### 数据结构的选择 为了降低时间复杂度并提高效率,可以引入`std::map`或者`unordered_map`来辅助实现去重功能[^1]。这些数据结构能够快速判断某项是否存在集合中,并支持高效的插入和查找操作。具体来说: - 使用 `std::set` 可以自动去除重复元素并对结果进行升序排列; - 如果还需要自定义比较逻辑,则可以选择基于哈希表的数据结构如 `unordered_set` 配合手动排序。 #### 排序策略 对于给定的一组DNA序列(通常表示为长度固定的字符串),按照字典顺序对其进行排序是一个常见需求。C++标准库提供了非常方便的方法来进行此类任务——即利用 `sort()` 函数配合合适的比较器函数对象或 lambda 表达式来指定所需的排序规则。 下面展示了一个简单的例子用于说明如何读取输入、执行必要的预处理步骤(包括但不限于删除冗余条目),最后输出经过整理的结果列表: ```cpp #include <bits/stdc++.h> using namespace std; int main(){ set<string> uniqueDNAs; string line, dna; while(getline(cin,line)){ stringstream ss(line); while(ss>>dna){ uniqueDNAs.insert(dna); // 自动过滤掉重复项 } } vector<string> sortedUnique(uniqueDNAs.begin(),uniqueDNAs.end()); sort(sortedUnique.begin(),sortedUnique.end()); for(auto it=sortedUnique.cbegin();it!=sortedUnique.cend();++it){ cout<<*it; if(next(it)!=sortedUnique.cend())cout<<" "; } } ``` 上述程序片段实现了基本的功能模块:从标准输入流逐行解析得到各个独立的DNA片段;借助 STL 容器特性轻松达成无重复记录维护目的;最终依据字母大小关系重新安排各成员位置后再统一打印出来[^3]。 #### 学习延伸至自然语言处理领域 值得注意的是,在计算机科学特别是机器学习方向上,“上下文”概念同样重要。例如 Word2Vec 这样的技术就是通过考察周围词语环境来捕捉特定词汇的意义特征[^2]。尽管两者应用场景差异显著,但从原理层面看均体现了对局部模式挖掘的关注。 ---
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