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POJ 1979 Red and Black 深度优先搜索上手题

2024-07-10 05:24:49   226人阅读

Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 21738 Accepted: 11656

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) 
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

Japan 2004 Domestic

题解

很简单的一道上手的深搜题目，题目意思很简单，就是不能走红色的砖。从起点开始深搜就行了。

代码示例

/*****************************************************************************
#       COPYRIGHT NOTICE
#       Copyright (c) 2014 All rights reserved
#       ----Stay Hungry Stay Foolish----
#
#       @author       :Shen
#       @name         :POJ 1979
#       @file         :G:\My Source Code\¡¾ACM¡¿ÑµÁ·\0630 - ËÑË÷\poj1979.cpp
#       @date         :2014/06/30 01:02
#       @algorithm    :DFS
******************************************************************************/

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

typedef long long int64;
int w, h, cnt;
char gird[25][25];
//driection u,  d,  r,  l;
int dx[4] = { 0,  0,  1, -1};
int dy[4] = {-1,  1,  0,  0};

inline bool inbound(int l, int r, int x)
{
    return (x >= l && x < r);
}

inline bool check(int x, int y)
{
    return inbound(0, w, x) && inbound(0, h, y);
}

void dfs(int x, int y)
{
    if (!check(x, y)) return;
    else if (gird[x][y] == '@' || gird[x][y] == '.')
    {
        cnt++;
        gird[x][y] = '#';
        for (int i = 0; i < 4; i++)
            dfs(x + dx[i], y + dy[i]);
    }
}

void solve()
{
    memset(gird, 0, sizeof(gird));
    int sx = 0, sy = 0; cnt = 0;
    for (int i = 0; i < h; i++)
    for (int j = 0; j < w; j++)
    {
        scanf(" %c", &gird[j][i]);
        if (gird[j][i] == '@') sx = j, sy = i;
    }
    dfs(sx, sy);
    printf("%d\n", cnt);
}

int main()
{
    while (scanf("%d%d", &w, &h))
    {
        if (w == 0 && h == 0) break;
        else solve();
    }
    return 0;
}


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