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            poj3083

            Children of the Candy Corn

            Time Limit: 1000MS Memory Limit: 65536K
            Total Submissions: 6055 Accepted: 2640

            Description

            The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.

            One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)

            As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.

            Input

            Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'.

            Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').

            You may assume that the maze exit is always reachable from the start point.

            Output

            For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.

            Sample Input

            2
            8 8
            ########
            #......#
            #.####.#
            #.####.#
            #.####.#
            #.####.#
            #...#..#
            #S#E####
            9 5
            #########
            #.#.#.#.#
            S.......E
            #.#.#.#.#
            #########

            Sample Output

            37 5 5
            17 17 9
            挺簡單的一個搜索題目,廣搜和深搜都要用
            題目意思,有一個迷宮,現在想知道總是沿著迷宮左側墻壁走的能走多少步,沿著迷宮右側墻壁走的能走多少步
            最少走多少步能出去
            我覺著這個題目關鍵在于如何處理怎樣沿著左右墻壁走
            我一開始以為設好總是右拐左拐的方向就行了,結果樣例就不過,最后輸出一看,有個地方一直來回走了,死循環了
            這里怎么處理呢?
            我也不會,網上看到神一般的代碼
             for (i=xx+1;i>=xx-2;i--)
            {
              j=(8+i)&3;//????
            }
            真心不明白啥意思,等弄懂了再回來寫思路
            除了這種方法,還沒想到別的方法,額,糾結了
              1#include<stdio.h>
              2#include<math.h>
              3#include<string.h>
              4#define MAX 45
              5#define MX  16000
              6struct node
              7{
              8    int x,y,d;
              9}
            ;
             10int w,h,sx,sy,ans1,ans2,ans3;
             11int dx[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
             12int q1;
             13char map[MAX][MAX];
             14short mark[MAX][MAX];
             15int inin(int xx,int yy)
             16{
             17    return xx>=0&&xx<h&&yy>=0&&yy<w;
             18}

             19void init()
             20{
             21    int i,j;
             22    int flag;
             23    flag=0;
             24    scanf("%d%d",&w,&h);
             25    for (i=0;i<h;i++)
             26    {
             27        scanf("%s",&map[i]);
             28        for (j=0;j<w;j++)
             29        if (map[i][j]=='S'&&!flag)
             30        {
             31            sx=i;sy=j;
             32            flag=1;
             33        }

             34    }

             35    for (i=0;i<4;i++)
             36    if (inin(sx+dx[i][0],sy+dx[i][1])&&map[sx+dx[i][0]][sy+dx[i][1]]=='.')
             37    {
             38        q1=i;break;
             39    }

             40}

             41int dfs1(int x,int y,int xx)
             42{
             43    int j,i,nx,ny;
             44    if (map[x][y]=='E')
             45    {
             46        return 1;
             47    }

             48    for (i=xx+1;i>=xx-2;i--)
             49    {
             50        j=(8+i)&3;//????
             51        nx=x+dx[j][0];ny=y+dx[j][1];
             52        if ((inin(nx,ny))&&(map[nx][ny]!='#'))
             53        {
             54            return 1+dfs1(nx,ny,j);
             55        }

             56    }

             57}

             58int dfs2(int x,int y,int xx)
             59{
             60    int j,i,nx,ny;
             61    if (map[x][y]=='E')
             62        return 1;
             63    for (i=xx-1;i<=xx+2;i++)
             64    {
             65        j=(8+i)&3;//????
             66        nx=x+dx[j][0];ny=y+dx[j][1];
             67        if ((inin(nx,ny))&&(map[nx][ny]!='#'))
             68        {
             69            return 1+dfs2(nx,ny,j);
             70        }

             71    }

             72}

             73int bfs()
             74{
             75    int i,j,head,tail,nx,ny;
             76    struct node q[MX],now;
             77    memset(mark,0,sizeof(mark));
             78    head=0;tail=1;
             79    mark[sx][sy]=1;
             80    q[tail].x=sx;q[tail].y=sy;q[tail].d=1;
             81    while (head<tail)
             82    {
             83        head++;
             84        now=q[head];
             85        for (i=0;i<4;i++)
             86        {
             87            nx=now.x+dx[i][0];
             88            ny=now.y+dx[i][1];
             89            if (inin(nx,ny)&&map[nx][ny]!='#'&&!mark[nx][ny])
             90            {
             91                if (map[nx][ny]=='E')
             92                {
             93                    return now.d+1;
             94                }

             95                tail++;
             96                q[tail].x=nx;q[tail].y=ny;q[tail].d=now.d+1;
             97                mark[nx][ny]=1;
             98            }

             99        }

            100    }

            101}

            102void work()
            103{
            104    int i,j;
            105    ans3=bfs();
            106    ans1=dfs1(sx,sy,q1);
            107    ans2=dfs2(sx,sy,q1);
            108}

            109int main()
            110{
            111    int t,ii;
            112    scanf("%d",&t);
            113    for (ii=1;ii<=t;ii++)
            114    {
            115        memset(map,0,sizeof(map));
            116        init(); 
            117        work();
            118        printf("%d %d %d\n",ans1,ans2,ans3);
            119    }

            120    return 0;    
            121}

            122

            posted on 2012-03-08 20:19 jh818012 閱讀(821) 評論(3)  編輯 收藏 引用

            評論

            # re: poj3083 2012-03-31 20:53 王私江

            額,呵呵,看看我的題解
            http://www.shnenglu.com/ArcTan/articles/169695.html
              回復  更多評論   

            # re: poj3083 2012-03-31 20:57 王私江

            嚓,位運算我還是不懂………………
            那個(8+i)&3應該就跟我那個循環去找一樣的哈。  回復  更多評論   

            # re: poj3083 2012-08-04 21:45 游客

            @王私江
            (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)   回復  更多評論   

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            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)
            • --游客
            • 3.?re: poj3414[未登錄]
            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評論內容較長,點擊標題查看
            • --王私江
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