Time limit1000 ms
Memory limit30000 kB
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13 题意：红色和黑色的地砖，只能走黑色的地砖，问最多可以走几块地砖题解：dfs
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
using namespace std;
#define PI 3.14159265358979323846264338327950
#define INF 0x3f3f3f3f3f3f3f3f;
char a[25][25];
int vis[25][25];
int m,n,st,en,sum;
void dfs(int x,int y)
{
a[x][y]=‘#‘;
sum++;
if(x-1>=0 && a[x-1][y]==‘.‘)
dfs(x-1,y);
if(x+1<n && a[x+1][y]==‘.‘)
dfs(x+1,y);
if(y-1>=0 && a[x][y-1]==‘.‘)
dfs(x,y-1);
if(y+1<m && a[x][y+1]==‘.‘)
dfs(x,y+1);
}
int main()
{
while(scanf("%d %d",&m,&n) && (m||n))
{
sum=1;
int i,j;
memset(vis,0,sizeof(vis));
for( i=0;i<n;i++)
for(j=0;j<m;j++)
{
cin>>a[i][j];
if(a[i][j]==‘@‘)
{
st=i;
en=j;
}
}
int x=st,y=en;
a[x][y]=‘#‘;
if(x-1>=0 && a[x-1][y]==‘.‘)
dfs(x-1,y);
if(x+1<n && a[x+1][y]==‘.‘)
dfs(x+1,y);
if(y-1>=0 && a[x][y-1]==‘.‘)
dfs(x,y-1);
if(y+1<m && a[x][y+1]==‘.‘)
dfs(x,y+1);
printf("%d\n",sum);
}
}
原文地址：https://www.cnblogs.com/smallhester/p/9499285.html