首页 > ACM题库 > HDU-杭电 > HDU 1312 Red and Black-DFS-[解题报告] C++
2013
12-09

HDU 1312 Red and Black-DFS-[解题报告] C++

Red and Black

问题描述 :

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.

输入:

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)

输出:

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).

样例输入:

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出:

45
59
6
13

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

char maps[25][25];
int r,c,sx,sy,res;
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};

void dfs(int a,int b)
{
    for(int i=0;i<4;i++)
    {
        int xx=a+dx[i];
        int yy=b+dy[i];
        if(xx>=1&&xx<=r&&yy>=1&&yy<=c&&maps[xx][yy]=='.')
        {
            maps[xx][yy]='#';
            res++;
			dfs(xx,yy);
        }
    }
}

int main()
{
    int i,j;
	while(scanf("%d%d",&c,&r)&&r&&c)
    {	
		memset(maps,0,sizeof(maps));
		res=1;
        getchar();
        for(i=1;i<=r;i++)
        {
            for(j=1;j<=c;j++)
            {
                scanf("%c",&maps[i][j]);
                if(maps[i][j]=='@')
                {
                    sx=i;
                    sy=j;
                    maps[i][j]='#';
                }
            }
            getchar();
        }
        dfs(sx,sy);
        printf("%d\n",res);
    }
    return 0;
}

解题报告转自:http://blog.csdn.net/u011262722/article/details/9614697


,
  1. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count

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