Red and Black
Time Limit:1000MSMemory Limit:30000K
Total Submissions:24058Accepted:13007
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.
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
4559613
题意：问从@出发可以到的点的个数
简单搜索！！！
参考代码：
#include <iostream>#include <string.h>#include <algorithm>using namespace std;typedef long long ll;char map[22][22];bool used[22][22];int dp[22][22];int n,m;int dirx[]={ 0,-1,0,1};int diry[]={-1, 0,1,0};int dfs(int x,int y){int sum=0;if (used[x][y]==true)return 0;if (dp[x][y]!=0)return dp[x][y];used[x][y]=true;for (int i=0;i<4;i++){int xx=x+dirx[i],yy=y+diry[i];if (map[xx][yy]=='.' && used[xx][yy]==false && xx<m && xx>=0 && yy<n && yy>=0){sum+=dfs(xx,yy);}}dp[x][y]=sum+1;return sum+1;}int main(){while (cin>>n>>m){if (n==0&&m==0)break;int x,y;memset(used,false,sizeof(used));memset(dp,0,sizeof(dp));for (int i=0;i<m;i++){for (int j=0;j<n;j++){cin>>map[i][j];if (map[i][j]=='@'){x=i;y=j;}}}cout<<dfs(x,y)<<endl;}return 0;}
，躲在某一时间想念一段时光的掌纹，
