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poj 1979 && zoj 2165 Red and Black

2024-07-17 10:57:37   227人阅读

Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 22409 Accepted: 12100

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

4559613

 

 

 

             一道简单的DFS题，统计可以搜索的个数。题意：一男子站在一个黑色的瓷砖上，他旁边有黑色和红色的瓷砖，但他只能走黑色的瓷砖，而且是4个方向，问他最多能走多少个黑色的瓷砖。

 

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int num;char map[25][25];int visited[25][25];int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0};void DFS(int x, int y, int n, int m){    int mx, my;    for(int i = 0; i<4; i++)    {        mx = x+dir[i][0];    my = y+dir[i][1];        if(mx>=1 && mx<=n && my>=1 && my<=m && !visited[mx][my] && map[mx][my]==‘.‘)        {            visited[mx][my] = 1;            num++;            DFS(mx, my, n, m);        }    }}int main(){    int n, m, x, y;    while(scanf("%d%d", &m, &n)!=EOF && (n || m))    {        num = 1;        memset(visited, 0, sizeof(visited));        for(int i = 1; i<=n; i++)        {            for(int j = 1; j<=m; j++)            {                cin>>map[i][j];                if(map[i][j] == ‘@‘)                {                    x = i;                    y = j;                    visited[x][y] = 1;                }            }        }        DFS(x, y, n, m);        printf("%d\n", num);    }    return 0;}

 


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