POJ-1979 Red and Bla

题目传送门
Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 37356 Accepted: 20315
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)
The end of the input is indicated by a line consisting of two zeros.
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

Japan 2004 Domestic

题目中有如下要求：
1.只能走周围的4个相邻点
2.只能走黑色点，不能走红色点
3.一次只能走一点

需要计算的是：能走到的黑色点的和

因为”.“表示黑色点，所以在下面的DFS函数中需要判断当前点为黑色点才可以进行下一步搜索。

并且下一步走的方向在于下方代码中定义的x[],y[]数组。

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m;char map[100][100];bool used[100][100]; int x[]={1,-1,0,0};int y[]={0,0,1,-1};int s_x,s_y;int node_cnt;void DFS(int node_x,int node_y){    if(node_x<0||node_x>=n||node_y<0||node_y>=m)    return ;    if(map[node_x][node_y]=='#')    return ;    node_cnt++;    used[node_x][node_y]=true;    for(int i=0;i<4;i++)    {        if(!used[node_x+x[i]][node_y+y[i]])        DFS(node_x+x[i],node_y+y[i]);    }}int main(){    while(scanf("%d%d",&m,&n)&&n&&m)    {        memset(used,false,sizeof(used));        getchar();        for(int i=0;i<n;i++)        {            scanf("%s",map[i]);            for(int j=0;j<m;j++)            if(map[i][j]=='@')            s_x=i,s_y=j;        }        node_cnt=0;        DFS(s_x,s_y);        printf("%d\n",node_cnt);    }    return 0;}