题目代号:HDU 1312

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20820    Accepted Submission(s): 12673

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

题目大意:一个男子站在‘.’上,他不能走到‘#’上问他能走到的‘.’的数量是多少。

解题思路:初始点bfs四个方向都遍历一次即可。

差点被自己气哭,第一次提交的时候忘记初始化数组,因为没有判断边界,直接导致WA。

AC代码:

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL; const int MAXM=;
char a[MAXM][MAXM];
int n,m,ans;
int cx[]={-,,,};
int cy[]={,,-,}; struct node
{
int x,y;
}; queue<node>Q; void bfs()
{
while(!Q.empty())
{
int x=Q.front().x;
int y=Q.front().y;
Q.pop();
for(int i=;i<;i++)
{
int tx=x+cx[i];
int ty=y+cy[i];
if(a[tx][ty]=='.')
{
ans++;
a[tx][ty]='#';
Q.push(node{tx,ty});
}
}
}
} int main()
{
//freopen("in.txt", "r", stdin);
while(cin>>m>>n,n&&m)
{
mem(a,);
for(int i=;i<=n;i++)
{
cin>>a[i]+;
for(int j=;j<=m;j++)
{
if(a[i][j]=='@')
{
Q.push(node{i,j});
a[i][j]='#';
}
}
}
ans=;
bfs();
cout<<ans<<endl;
}
return ;
}

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