杭电1312 深搜和广搜

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

题意 ：

点的可以经过 #的不可以经过 @为起点

巧妙的是可以把.转换成# 当访问过之后

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;
};
bool YES(int x0,int y0) { //xx,yy

    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]='#';//将.转换成# 
                sum++;
                q.push(endd);
            }
        }
    }
}
int main() {
    int i,j,di,dj;
    while (scanf("%d%d",&x,&y)!=EOF&&x&&y) {
        getchar();
        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;
}

 

 

 

深搜

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[25][25];
int vis[25][25];
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int w,h;
int dfs(int si,int sj)
{
    map[si][sj]='#';
    vis[si][sj]=1;
    for(int i=0;i<4;i++)
    {
        int x=si+dir[i][0];
        int y=sj+dir[i][1];
        if(x<0||x>=h||y<0||y>=w||map[x][y]=='#')
            continue;
        dfs(x,y);
    }
}
int  main()
{
    while(~scanf("%d%d",&w,&h)&&w&&h)
    {
        memset(vis,0,sizeof(vis));
        for(int i=0;i<h;i++)
            scanf("%s",map[i]);
        bool flag=false;
        for(int i=0;i<h;i++)
        {
            for(int j=0;j<w;j++)
            {
                if(map[i][j]=='@')
                {
                    dfs(i,j);
                    flag=true;
                    break;
                }
            }
            if(flag)
                break;
        }
        int num=0;
        for(int i=0;i<h;i++)
            for(int j=0;j<w;j++)
            {
                if(vis[i][j])
                    num++;
            }
        printf("%d\n",num);
    }
    return 0;
}