广度优先搜素和深度优先搜素

以北大的1979为例：

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17144		Accepted: 9025

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

具体要求：

输入相应的格子的数目，并且标记每个格子的颜色，当遇到黑色的格子时可以踩上，当遇到红色的格子时不能踩上，要求输出从一个起始点出发能够踩到的黑色格子数。

 

算法思想：

1，  可以根据图的深度优先搜索，计算并输出通过深度搜索所通过的节点数，即相应的格子数目。

2，  可以根据图的广度优先搜索，计算并输出通过广度搜索所通过的节点数，即相应的格子数目。

 

算法实现：

1，  深度优先搜索：通过递归方法，从图结构的一个结点开始深度搜索。相应的代码如下

#include<stdio.h> #include<string.h>

int a,b,n,m; char c[22][22]; int sert(int a,int b) { if(c[a][b]=='#'||a>=m||b>=n||a<0||b<0) return 0; else { c[a][b]='#'; return 1+sert(a,b+1)+sert(a,b-1)+sert(a-1,b)+sert(a+1,b); } } int main() { int i,j; while(scanf("%d%d",&n,&m),n||m) { for(i=0;i<m;i++) { getchar(); for(j=0;j<n;j++) { scanf("%c",&c[i][j]); if(c[i][j]=='@') { a=i;b=j; } } } n=sert(a,b); printf("%d\n",n); } return 0; }

广度优先搜索：通过建立队列实现广度搜索，每走到符合要求的格子时，当前格子进队列。相应的代码如下：

#include<stdio.h> #include<iostream> #include<queue>

using namespace std; typedef struct { int x,y; }node; int n,m,a,b; char c[25][25]; void sert() { int i,f[4][2]={0,1,0,-1,1,0,-1,0},p=0;; queue<node> q; node t,temp; t.x=a; t.y=b; q.push(t); while(!q.empty()) { t=q.front(); q.pop(); for(i=0;i<4;i++) { temp.x=t.x+f[i][0]; temp.y=t.y+f[i][1]; if(temp.x>=0&&temp.x<m&&temp.y<n&&temp.y>=0&&c[temp.x][temp.y]!='#') { p++; q.push(temp); c[temp.x][temp.y]='#'; } } } printf("%d\n",p); } int main() { int i,j; while(scanf("%d%d",&n,&m),n||m) { for(i=0;i<m;i++) { getchar(); for(j=0;j<n;j++) { scanf("%c",&c[i][j]); if(c[i][j]=='@') { a=i; b=j; } } } sert(); } return 0; }