HDU 1312 Red and Black(经典搜索,DFS&BFS三种方式)

Red and Black

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 

 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 

 

Sample Input

  

   6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
  

 

Sample Output

  

   45
59
6
13
  

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312

题意:根据地图,问可以到达多少地方.'@'为起点, ' . '为路,可以到达, ' # '为墙,不能通过.

思路:简单题,既可以用深搜,也可以用广搜,广搜是可以用C++里的queue,如果是C的话也可以实现其功能,其实也没有多复杂.

注意:输入的两个数,第一个是列数,第二个是行数!!!

还有一个小技巧,搜索过的地方可以用'#'标记!

AC代码1:经典深搜:

#include<iostream>
#include <cstdio>//getchar()头文件
using namespace std;
char a[25][25];
int dir[4][2]=
{
    {1,0}, //向右
    {-1,0},//向左
    {0,1}, //向上
    {0,-1} //向下
};
int x,y,num;
bool YES(int x0,int y0)//xx,yy
{
   //x--列,y--行
    if(x0<y&&x0>=0&&y0>=0&&y0<x)
        return true;
    else
        return false;
}
void DFS(int x0,int y0)
{
    a[x0][y0]='#';
    num++;
    int xx,yy;
    for(int i=0; i<4; i++)
    {
        int xx=x0+dir[i][0];
        int yy=y0+dir[i][1];
        if(YES(xx,yy)&&a[xx][yy]=='.')
            DFS(xx,yy);
    }

}
int main()
{
    int i,j,di,dj;
    while (scanf("%d%d",&x,&y)!=EOF)
    {
        getchar();
        if (x==0&&y==0) break;
        for (i=0; i<y; i++)
        {
            for (j=0; j<x; j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j] == '@')
                {
                    di = i;
                    dj = j;
                }
            }
            getchar();
        }
        num=0;
        DFS(di,dj);
        cout<<num<<endl;

    }
    return 0;
}

AC代码2,经典广搜:

#include<cstdio>
#include<queue>
using namespace std;
char a[25][25];
int x,y,sum;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
struct node
{
    int x,y;
}q[500];
bool YES(int x0,int y0)//xx,yy
{
   //x--列,y--行
    if(x0<y&&x0>=0&&y0>=0&&y0<x)
        return true;
    else
        return false;
}
void BFS(int x0,int y0)
{
    queue<node>q;
    node start,endd;
    start.x=x0;
    start.y=y0;
    q.push(start);
    while(!q.empty())
    {
        start=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            endd.x=start.x+dir[i][0];
            endd.y=start.y+dir[i][1];
            if(YES(endd.x,endd.y)&&a[endd.x][endd.y]=='.')
            {
                a[endd.x][endd.y]='#';
                sum++;
                q.push(endd);
            }
        }
    }
}
int main()
{
    int i,j,di,dj;
    while (scanf("%d%d",&x,&y)!=EOF)
    {
        getchar();
        if (x==0&&y==0) break;
        for (i=0; i<y; i++)
        {
            for (j=0; j<x; j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j] == '@')
                {
                    di = i;
                    dj = j;
                }
            }
            getchar();
        }
        sum=1;
        BFS(di,dj);
        printf("%d\n",sum);
    }
    return 0;
}

AC代码3:C语言广搜.

#include<cstdio>
#include<queue>
using namespace std;
int a[25][25];
int x,y,sum;
int dir[4][2]={1,0,0,1,-1,0,0,-1};
struct node
{
    int x,y;
}q[500];
bool YES(int x0,int y0)//xx,yy
{
   //x--列,y--行
    if(x0<y&&x0>=0&&y0>=0&&y0<x)
        return true;
    else
        return false;
}
void BFS(int x0,int y0)
{
    int head=0,tail=1;
    q[1].x=x0;
    q[1].y=y0;
    while(head<tail)
    {
        ++head;
        for(int i=0;i<4;i++)
        {
            int x1=q[head].x+dir[i][0];
            int y1=q[head].y+dir[i][1];
            if(YES(x1,y1)&&a[x1][y1]=='.')
            {
                a[x1][y1]='#';
                sum++;
                q[++tail].x=x1;
                q[tail].y=y1;
            }
        }
    }
}
int main()
{
    int i,j,di,dj;
    while (scanf("%d%d",&x,&y)!=EOF)
    {
        getchar();
        if (x==0&&y==0) break;
        for (i=0; i<y; i++)
        {
            for (j=0; j<x; j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j] == '@')
                {
                    di = i;
                    dj = j;
                }
            }
            getchar();
        }
        sum=1;
        BFS(di,dj);
        printf("%d\n",sum);
    }
    return 0;
}