【杭电】[1312]Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

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

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感觉做过好多遍的题都没贴-.- 最简单的DFS迷宫搜索问题 递归查找能走的路径

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#include<stdio.h>
char map[22][22];
int move[6]= {1,-1,0,0};
int W,H,cnt;
void dfs(int n,int m) {
    cnt++;
    map[n][m]='#';
    for(int i=0; i<4; i++) {
        int tn=n+move[i],tm=m+move[(i+2)%4];
        if(tn>=0&&tn<H&&tm>=0&&tm<W&&map[tn][tm]!='#')
            dfs(tn,tm);
    }
    return ;
}
int main() {
    while(scanf("%d %d",&W,&H),W&&H) {
        getchar();
        int mH,mW;
        for(int i=0; i<H; i++) {
            for(int j=0; j<W; j++) {
                map[i][j]=getchar();
                if(map[i][j]=='@')
                    mH=i,mW=j;
            }
            getchar();
        }
        cnt=0;
        dfs(mH,mW);
        printf("%d
",cnt);
    }
    return 0;
}

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题目地址:【杭电】[1312]Red and Black

查看原文：http://www.boiltask.com/blog/?p=1915