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hdu 1312 Red and Black

时间:2015-12-10 15:07:06      阅读:135      评论:0      收藏:0      [点我收藏+]

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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. 
 

 

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output
45
59
6
13
 
Source

 

题解: 求 “.”数,  注意  行和列  吧...     找了好久才找到错在哪里...

 

代码:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <math.h>
 4 #include <limits.h> 
 5 #include <algorithm>
 6 #include <iostream>
 7 #include <ctype.h>
 8 #include <iomanip>
 9 #include <queue>
10 #include <map>
11 #include <stdlib.h>
12 using namespace std;
13 
14 int ans,m,n;
15 char v[30][30];
16 int dx[4]={0,0,1,-1};
17 int dy[4]={1,-1,0,0};
18 
19 void dfs(int x,int y)
20 {
21     ans++;
22     v[x][y]=#;
23     for(int k=0;k<4;k++){
24         int px=x+dx[k]; 
25         int py=y+dy[k]; 
26         if(v[px][py]==. && py<m && px<n && px>=0 && py>=0){ 
27             dfs(px,py);
28         }
29     }
30 }
31 
32 int main()
33 {
34     while(~scanf("%d%d",&m,&n)&&m!=0&&n!=0){
35         int e,d,i,j;
36         for(i=0;i<n;i++)
37             scanf("%s",v[i]);
38         for(i=0;i<n;i++){
39             for(j=0;j<m;j++){
40                 if(v[i][j]==@){
41                     e=i;
42                     d=j;
43                 }
44             }
45         }
46         ans=0;
47         dfs(e,d); 
48         printf("%d\n",ans);
49     }
50 }

 

 

hdu 1312 Red and Black

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原文地址:http://www.cnblogs.com/wangmengmeng/p/5035895.html

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