题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1312
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10724 Accepted Submission(s): 6716
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
Asia 2004, Ehime (Japan), Japan Domestic
题意: 在一个图中给出起点以及障碍物，求从该点出发所能到达的点的个数。
题解: 简单深搜，都不用任何剪枝——走过的点标记一下就好~~ 直接递归
AC代码:
#include<iostream>
#define maxn 25
using namespace std;
char chess[maxn][maxn],ch;
int sx,sy,w,h;
int dfs(int x,int y){
int sum=0;
if(x<0||x>=h||y<0||y>=w||chess[x][y]=='#')return 0;
chess[x][y]='#';
return 1+dfs(x+1,y)+dfs(x,y+1)+dfs(x-1,y)+dfs(x,y-1);
}
int main()
{
while(cin>>w>>h&&(w||h))
{
for(int i=0;i<h;i++)
for(int j=0;j<w;j++){
cin>>ch; chess[i][j]=ch;
if(ch=='@'){
chess[i][j]='.'; sx=i; sy=j;
}
}
cout<<dfs(sx,sy)<<endl;
}
}