ZOJ-2165(DFS)
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Red and Black
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 67 Accepted Submission(s) : 49
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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.
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)
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.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0
Sample Output
4559613
Source
Asia 2004, Ehime (Japan), Japan Domestic
连通块问题,
当初怎么改都没改出来,
突然发现是先列后行。。
思路与最大黑区域相同:
#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int Map[21][21];char s[21][21];int n,m;int ax,ay;int cnt;int dir[4][2]={1,0,-1,0,0,1,0,-1};int mx;int my;void dfs(int x,int y){ for(int i=0;i<4;i++) { mx=x+dir[i][0]; my=y+dir[i][1]; if(Map[mx][my]==1&&mx>=0&&my>=0&&mx<n&&my<m) { cnt++; Map[mx][my]=0; dfs(mx,my); } }}int main(){ while(~scanf("%d%d",&m,&n)&&n&&m) { cnt=1; memset(Map,0,sizeof(Map)); for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { cin>>s[i][j]; if(s[i][j]=='@') { ax=i; ay=j; } if(s[i][j]=='.') { Map[i][j]=1; } } } Map[ax][ay]=0; dfs(ax,ay); cout<<cnt<<endl; } return 0;}
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