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原题目：
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.
OutputFor 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 0Sample Output
45 59 6 13 -
题目大意：
给出一个w列和h行的方阵，方阵中有’.’,’#‘和’@’。其中’.‘是可以移动的地方，’#‘是红色的砖，是不可以移动的地 方，’@'是最初人的地点就（图的入口），人可以上下左右移动，以0 0 为结束。
要求输出人可以移动的砖的个数。该题利用搜索。
注意：
先输入w，然后输入h，但是w是列数，h是行数。
- 代码：
#include<iostream>
#include<cstring>
using namespace std;
void dfs(int x,int y);
int w,h,tmp=0;
char ma[100][100];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int main(){
while(cin>>h>>w){
if(w==0&&h==0) break;
memset(ma,'0',sizeof(ma));
int ax,ay;
for(int i=0;i<w;i++){
for(int j=0;j<h;j++){
cin>>ma[i][j];
if(ma[i][j]=='@'){
ax=i;
ay=j;
}
}
}
tmp=0;
dfs(ax,ay);
cout<<tmp<<endl;
}
return 0;
}
void dfs(int x,int y){
ma[x][y]='#';
tmp++;
for(int i=0;i<4;i++){
int nx=x+dx[i];
int ny=y+dy[i];
if(ma[nx][ny]=='.'&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]=='.'){
dfs(nx,ny);
}
}
}
Select Code
#include<iostream>
#include<cstring>
using namespace std;
void dfs(int x,int y);
int w,h,tmp=0;
char ma[100][100];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int main(){
while(cin>>h>>w){
if(w==0&&h==0) break;
memset(ma,'0',sizeof(ma));
int ax,ay;
for(int i=0;i<w;i++){
for(int j=0;j<h;j++){
cin>>ma[i][j];
if(ma[i][j]=='@'){
ax=i;
ay=j;
}
}
}
tmp=0;
dfs(ax,ay);
cout<<tmp<<endl;
}
return 0;
}
void dfs(int x,int y){
ma[x][y]='#';
tmp++;
for(int i=0;i<4;i++){
int nx=x+dx[i];
int ny=y+dy[i];
if(ma[nx][ny]=='.'&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]=='.'){
dfs(nx,ny);
}
}
}
POJ上不支持
bits/stdc++的头文件，很尴尬。