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.
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 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)
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13题目的大意就是搜图判断最多能遍历多少个".";简单dfs
#include<iostream>
#include<cstring>
using namespace std;
int w,h,ans;
char arr[25][25];
int mark[25][25];
int d[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
void dfs(int x,int y)
{
mark[x][y]=1;
for(int i=0;i<4;i++){
int dx=x+d[i][0];
int dy=y+d[i][1];
if(dx>=0&&dy>=0&&dx<h&&dy<w&&mark[dx][dy]==0&&arr[dx][dy]==‘.‘){
ans++;
mark[dx][dy]=1;
dfs(dx,dy);
}
}
}
int main()
{
while(cin>>w>>h){
memset(mark,0,sizeof(mark));
ans=1;
if(w==0&&h==0)
break;
for(int i=0;i<h;i++){
scanf("%s",&arr[i]);
}
for(int i=0;i<h;i++)
for(int j=0;j<w;j++){
if(arr[i][j]==‘@‘){
// mark[i][j]=1;
dfs(i,j);
}
}
cout<<ans<<endl;
}
return 0;
}
BFS也可以写就是只要是x周围存在"."就加1,
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
char arr[22][22];
int sa,ea;
int n,m;
int v[22][22]={0};
struct stu{
int a,b;
// int sum;
};
int a[4]={0,1,0,-1};
int b[4]={1,0,-1,0};
void bfs()
{
int ans=0;
// priority_queue<stu >que;
queue<stu>que;
stu q1;
q1.a=sa;
q1.b=ea;
// q1.sum=1;
v[sa][ea]=1;
que.push(q1);
while(que.size()){
stu h;
// h=que.top();
h=que.front();
que.pop();
stu d;
for(int i=0;i<4;i++){
d.a=h.a+a[i];
d.b=h.b+b[i];
if(d.a>=0 && d.b>=0 && d.a<m && d.b<n&& v[d.a][d.b]!=1 && arr[d.a][d.b]!=‘#‘){
v[d.a][d.b]=1;
// d.sum=h.sum+1;
// cout<<d.sum<<endl;
que.push(d);
// ans=max(ans,d.sum);
ans++;//只要加入到队列中就加一，因为加入到队列的一定是‘.‘
}
}
}
cout<<ans+1<<endl;
}
int main(){
while(cin>>n>>m)
{
memset(v,0,sizeof(v));
if(n==0&&m==0)
break;
for(int i=0;i<m;i++){
scanf("%s",&arr[i]);
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(arr[i][j]==‘@‘){
sa=i;
ea=j;
}
}
}
bfs();
}
return 0;
}
原文地址：https://www.cnblogs.com/Accepting/p/11241635.html