hdu1312 Red and Black

标签：multi   tput   避免   The   stream   print   HERE   out   nts   

题目链接

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

分析：

搜索模板题。BFS和DFS应该都可以，不过BFS比较快。

程序说明：

int nxt[][]：方向数组。

queue<pair<int,int> > que：BFS的队列，用于扩展点。其中，pair的第一维为x，第二维为y。（也就是坐标）

bool book[]：标记数组，避免搜索中重复扩展。

code:

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int ans=1,n,m,startx,starty,nxt[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
char a[30][30];
bool book[30][30];
void bfs()
{
    queue<pair<int,int> > que; 
    que.push(make_pair(startx,starty));
    book[startx][starty]=true;
    while(!que.empty())
    {
        for(int k=0;k<4;k++)
        {
            int tx=que.front().first+nxt[k][0];
            int ty=que.front().second+nxt[k][1];
            if(tx<0||tx>=n||ty<0||ty>=m) continue;
            if(book[tx][ty]==false&&a[tx][ty]=='.')
            {
                book[tx][ty]=true;
                ans++;
                que.push(make_pair(tx,ty));
            }
        }
        que.pop();
    }
}
int main()
{
    while(~scanf("%d%d",&m,&n)&&(m||n))
    {
        memset(a,0,sizeof(a));
        memset(book,false,sizeof(book));
        ans=1;
        startx=starty=0;
        int flag=0;
        for(int i=0;i<n;i++) 
        {
            getchar();
            for(int j=0;j<m;j++) 
            {
                a[i][j]=getchar();
                if(a[i][j]=='@') startx=i,starty=j;
            } 
        }
        bfs();
        printf("%d\n",ans);
    }
    return 0;
}

hdu1312 Red and Black

标签：multi   tput   避免   The   stream   print   HERE   out   nts   

原文地址：https://www.cnblogs.com/juruo-zzt/p/12247943.html