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POJ1979 Red and Black

时间:2016-06-14 23:44:01      阅读:180      评论:0      收藏:0      [点我收藏+]

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速刷一道DFS

 

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. 
@是起点,#不能走,‘.‘是正常道路,问从起点能到达多少格子

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

45
59
6
13

Source

 

 

刷道普通的DFS,复习模板

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=50;
 9 char mp[mxn][mxn];
10 int ans;
11 int W,H;
12 void dfs(int x,int y){
13     if(x<1 || x>H || y<1 || y>W)return;
14     if(mp[x][y]==#)return;
15     ans++;
16     mp[x][y]=#;
17     dfs(x+1,y);
18     dfs(x,y+1);
19     dfs(x-1,y);
20     dfs(x,y-1);
21     return;
22 }
23 int main(){
24     while(scanf("%d%d",&W,&H) && W && H){
25         memset(mp, ,sizeof(mp));
26         ans=0;
27         int i,j;
28         for(i=1;i<=H;i++){
29             scanf("%s",mp[i]+1);
30         }
31         int sx,sy;
32         bool flag=0;
33         for(i=1;i<=H;i++){
34               for(j=1;j<=W;j++)
35                   if(mp[i][j]==@){
36                       sx=i;sy=j;
37                       flag=1;
38                       break;
39                   }
40             if(flag)break;
41         }
42         dfs(sx,sy);
43         printf("%d\n",ans);
44     }
45     return 0;
46 }

 

POJ1979 Red and Black

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原文地址:http://www.cnblogs.com/SilverNebula/p/5585837.html

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