Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25283 Accepted Submission(s): 15239
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
题意：给一张图（注意n和m位置反了）从@点出发，“.”可以走，“#”不能走，问最多可以走多少个点
广搜深搜都可以写，这道题才是入门题，毕竟搜索是要先遍历的，而又得题说是入门题但是直接就要求最长走多少什么的有点过分。让很多新手曲解搜索的意义
广搜代码：
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
int a,b;
char mapp[50][50];
int vis[50][50];
struct node
{
int x,y;
};
node st,ed;
int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int judge(int x,int y)
{
if(x>=0&&x<a&&y>=0&&y<b&&mapp[x][y]=='.'&&vis[x][y]==0)
return 1;
return 0;
}
int ans;
int bfs()
{
queue<node>q;
while(!q.empty()) q.pop();
memset(vis,0,sizeof(vis));
vis[st.x][st.y]=1;
q.push(st);
while(!q.empty())
{
// printf("**\n");
st=q.front();
q.pop();
for(int i=0;i<4;i++)
{
ed.x=st.x+dx[i];
ed.y=st.y+dy[i];
if(judge(ed.x,ed.y))
{
vis[ed.x][ed.y]=1;
ans++;
q.push(ed);
}
}
}
return ans;
}
int main()
{
while(~scanf("%d%d",&b,&a))
{
if(a==0&&b==0)
break;
for(int i=0;i<a;i++)
scanf("%s",mapp[i]);
for(int i=0;i<a;i++)
for(int j=0;j<b;j++)
{
if(mapp[i][j]=='@')
{
st.x=i;
st.y=j;
}
}
ans=0;
// printf("%d %d\n",st.x,st.y);
printf("%d\n",bfs()+1);
}
}深搜代码（改编自广搜，所以结构体其实不必要开）
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
int a,b;
char mapp[50][50];
int vis[50][50];
struct node
{
int x,y;
};
node st,ed;
int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int judge(int x,int y)
{
if(x>=0&&x<a&&y>=0&&y<b&&mapp[x][y]=='.'&&vis[x][y]==0)
return 1;
return 0;
}
int ans;
void dfs(int x,int y)
{
vis[x][y]=1;
ans++;
for(int i=0;i<4;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(judge(xx,yy))
{
dfs(xx,yy);
}
}
return ;
}
int main()
{
while(~scanf("%d%d",&b,&a))
{
if(a==0&&b==0)
break;
for(int i=0;i<a;i++)
scanf("%s",mapp[i]);
for(int i=0;i<a;i++)
for(int j=0;j<b;j++)
{
if(mapp[i][j]=='@')
{
st.x=i;
st.y=j;
}
}
ans=0;
memset(vis,0,sizeof(vis));
dfs(st.x,st.y);
printf("%d\n",ans);
}
}