hdu 1312(DFS)

Posted 2020-08-20 见字如面

               Red and Black

Tme Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18126    Accepted Submission(s): 11045

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

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#[email protected]#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

[email protected]

###.###

..#.#..

..#.#..

0 0

 Sample Output

45

59

6

13

 题意就是  找从@点开始出发，问所能到达的‘.‘点的个数（最开始理解错了意思，以为是一条路径的最大值，后面

才发现是可以到达的‘.‘点 ），直接DFS，

#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstdlib>
#include<vector>
#include<set>
#include<queue>
#include<cstring>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N = 1e6+10;
const ll mod = 1e9+7;
int t=0,flag=0;
char s[26][26];
int visited[26][26];
int m,n;
int _x[4]={0,1,-1,0};
int _y[4]={1,0,0,-1};
int ans;
void DFS(int x,int y){
    for(int t=0;t<4;t++){
        int i=x+_x[t];
        int j=y+_y[t];
        if(i>=0&&j>=0&&visited[i][j]==0&&s[i][j]==‘.‘&&i<=n-1&&j<=m-1)
        {
            ans++;
            visited[i][j]=1;
            DFS(i,j);
        }
    }
}
int main(){
    while(scanf("%d%d",&m,&n)!=EOF){
        if(m==0&&n==0)break;
        memset(visited,0,sizeof(visited));
        int ii,jj;
        int i,j;
        for(ii=0;ii<n;ii++){
            for(jj=0;jj<m;jj++){
                cin>>s[ii][jj];
                if(s[ii][jj]==‘@‘){i=ii;j=jj;}
            }
        }
        ans=1;
        visited[i][j]=1;
        DFS(i,j);
        cout<<ans<<endl;
    }
}