农场灌溉问题ZOJ2412

描述:

一农场由图所示的十一种小方块组成,蓝色线条为灌溉渠。若相邻两块的灌溉渠相连则只需一口水井灌溉。

输入:

给出若干由字母表示的最大不超过50×50具体由(m,n)表示,的农场图

输出:

编程求出最小需要打的井数。每个测例的输出占一行。当M=N=-1时结束程序。

输入样例:

2 2 DK HF 3 3 ADC FJK IHE -1 -1

输出样例:

2 3


#include <iostream>

using namespace std;
int m, n;

struct Maze
{
    char ch;                     //农场类型
    int visited;                 //访问标记
    int left, right, up, down;   //连通线
}maze[100][100];
int dir[4][2] = {{1,0}, {-1,0}, {0,-1}, {0,1}}; //上下左右四个方向

bool Place(int x, int y, int nx, int ny)
{
    //未越界且没有被访问过
    if(nx>0 && nx<=m && ny>0 && ny<=n && !maze[nx][ny].visited)
    {
        //向左
        if(nx==x && ny==y-1 && maze[x][y].left && maze[nx][ny].right)
            return true;
        //向下
        if(nx==x+1 && ny==y && maze[x][y].down && maze[nx][ny].up)
            return true;
        //向右
        if(nx==x && ny==y+1 && maze[x][y].right && maze[nx][ny].left)
            return true;
        //向上
        if(nx==x-1 && ny==y && maze[x][y].up && maze[nx][ny].down)
            return true;
    }
    return 0;
}
void dfs(int x, int y)
{
    maze[x][y].visited = 1;              //当前点已访问
    for(int i = 0; i < 4; i++)
    {
        int nx = x+dir[i][0];
        int ny = y+dir[i][1];
        //将所有可连通的区域全部标记
        if(Place(x, y, nx, ny))
            dfs(nx, ny);
    }
}
int main()
{
    while(cin >> m >> n && (m!=-1 || n!=-1))
    {
        int cnt = 0;
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++)
            {
                cin >> maze[i][j].ch;
                //初始化
                maze[i][j].visited = 0;
                maze[i][j].down = 0;
                maze[i][j].up = 0;
                maze[i][j].right = 0;
                maze[i][j].left = 0;
                switch(maze[i][j].ch)
                {
                    case'A':    maze[i][j].left=1;
                                maze[i][j].up=1;
                                break;
                    case'B':    maze[i][j].right=1;
                                maze[i][j].up=1;
                                break;
                    case'C':    maze[i][j].left=1;
                                maze[i][j].down=1;
                                break;
                    case'D':    maze[i][j].right=1;
                                maze[i][j].down=1;
                                break;
                    case'E':    maze[i][j].up=1;
                                maze[i][j].down=1;
                                break;
                    case'F':    maze[i][j].left=1;
                                maze[i][j].right=1;
                                break;
                    case'G':    maze[i][j].left=1;
                                maze[i][j].right=1;
                                maze[i][j].up=1;
                                break;
                    case'H':    maze[i][j].left=1;
                                maze[i][j].up=1;
                                maze[i][j].down=1;
                                break;
                    case'I':    maze[i][j].left=1;
                                maze[i][j].right=1;
                                maze[i][j].down=1;
                                break;
                    case'J':    maze[i][j].right=1;
                                maze[i][j].up=1;
                                maze[i][j].down=1;
                                break;
                    case'K':    maze[i][j].left=1;
                                maze[i][j].right=1;
                                maze[i][j].up=1;
                                maze[i][j].down=1;
                                break;
                }
            }
        //连同区域的个数
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= n; j++)
            {
                if(maze[i][j].visited == 0)     //未被访问过
                {
                    dfs(i, j);
                    cnt ++;
                }
            }
        cout << cnt << endl;
    }
    return 0;
}



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