poj 1979DFS&&BFS

DescriptionThere 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.InputThe input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are thenumbers 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.汉语意思：一个有障碍的地图，给出一个起始点，可以走多少地图方块；@表示现在所在的位置，只能上下左右四个方向走，只能在‘.’这个上面走，#是障碍物输入：

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 06 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

输出：45
59
6
13

分析用的是深度搜索的知识

#include <iostream>
#include <string.h>
#define Size 30
using namespace std;
char Map[Size][Size];
bool visited[Size][Size];
int W,H;
int result;
int sx,sy;
int xmove[4]={1,0,-1,0};
int ymove[4]={0,1,0,-1};
int ok(int x,int y)
{
if(x>=0&&x<W&&y>=0&&y<H)
return 1;
return 0;
}
void DFS(int x,int y)
{
for(int i=0;i<4;i++)
{
int a=x+xmove[i];
int b=y+ymove[i];
if(ok(a,b))
{
if(!visited[a][b]&&Map[a][b]=='.')
{
visited[a][b]=1;
result++;
DFS(a,b);
}
}
}
}

int main()
{
while(cin>>H>>W)
{
if(!W&&!H)
break;
for(int i=0;i<W;i++)
{
for(int j=0;j<H;j++)
{
cin>>Map[i][j];
if(Map[i][j]=='@')
{
sx=i;
sy=j;
}
}
}
result=1;
memset(visited,0,sizeof(visited));
DFS(sx,sy);
cout<<result<<endl;
}
return 0;
}

BFS

<pre name="code" class="cpp">#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>using namespace std;#define maxn 25struct XPoint{int x, y;} a;bool map[maxn][maxn], vis[maxn][maxn];int n, m, ans;int dir[4][2] ={{ 0, 1 },{ 0, -1 },{ 1, 0 },{ -1, 0 } };void input(){for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){char ch;scanf("%c", &ch);if (ch == '#')map[i][j] = false;elsemap[i][j] = true;if (ch == '@'){a.x = i;a.y = j;}}getchar();}}bool ok(XPoint &a){if (a.x < 0 || a.y < 0 || a.x >= n || a.y >= m)return false;return !vis[a.x][a.y] && map[a.x][a.y];}void bfs(){memset(vis, 0, sizeof(vis));queue<XPoint> q;q.push(a);vis[a.x][a.y] = true;ans = 0;while (!q.empty()){XPoint a = q.front();q.pop();ans++;XPoint b;for (int i = 0; i < 4; i++){b.x = a.x + dir[i][0];b.y = a.y + dir[i][1];if (ok(b)){q.push(b);vis[b.x][b.y] = true;}}}}int main(){//freopen("t.txt", "r", stdin);while (scanf("%d%d", &m, &n), m | n){getchar();input();bfs();printf("%d\n", ans);}return 0;}

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