Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19690 Accepted Submission(s): 11965
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
Sample Output
45
59
6
13
好久之前老大讲过的,忘了写了。。。
#include<bits/stdc++.h> using namespace std; int direct [4][2]={-1,0,1,0,0,1,0,-1}; //方向 char str[25][25]; bool flag[25][25]; int w,h,ans; void DFS(int x,int y){ for(int i=0;i<4;i++){ int p=x+direct[i][0]; int q=y+direct[i][1]; if(p>=0&&q>=0&&p<h&&q<w&&flag[p][q]==0&&str[p][q]==‘.‘){ //判断,至今仍不明白方向为-1的时候怎么还是>=0,还是我太蠢了。。。 ans++; flag[p][q]=1; DFS(p,q); //递归 } } } int main(){ int i,j,k; int Dx,Dy; while(~scanf("%d%d",&w,&h)){ if(w==0&&h==0)break; memset(flag,0,sizeof(flag)); //这个函数通常为新申请的内存做初始化工作 getchar(); for(i=0;i<h;i++){ for(j=0;j<w;j++){ scanf("%c",&str[i][j]); if(str[i][j]==‘@‘){ Dx=i; Dy=j; } } getchar(); } ans=1; flag[Dx][Dy]=1; DFS(Dx,Dy); printf("%d\n",ans); } return 0; }
写下来,等忘的时候再回来看。
啊,我这个菜鸟,好笨。。。┭┮﹏┭┮