• NUC1397 Oil Deposits【DFS】


    Oil Deposits

    时间限制: 1000ms 内存限制: 65535KB

    问题描述
    The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
    输入描述
    The input contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.
    输出描述
    are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
    样例输入
    1 1
    *
    3 5
    *@*@*
    **@**
    *@*@*
    1 8
    @@****@*
    5 5 
    ****@
    *@@*@
    *@**@
    @@@*@
    @@**@
    0 0
    样例输出
    0
    1
    2
    2
    
    来源
    Mid-Central USA 1997



    问题分析:

    这个问题和《HDU1241 POJ1562 UVa572 UVALive5317 Oil Deposits》是同一个问题,代码拿过来用就AC了。

    程序说明:

    参见参考链接。

    参考链接:HDU1241 POJ1562 UVa572 UVALive5317 Oil Deposits

    题记:

    程序做多了,不定哪天遇见似曾相识的。


    AC的C++程序如下:

    /* HDU1241 POJ1562 UVa572 UVALive5317 Oil Deposits */
    
    #include <stdio.h>
    #include <memory.h>
    
    #define DIRECTSIZE 8
    
    struct direct {
        int drow;
        int dcol;
    } direct[DIRECTSIZE] =
        {{0, -1}, {0, 1}, {-1, 0}, {1, 0}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
    
    #define MAXN 100
    
    char grid[MAXN+2][MAXN+2];
    
    void dfs(int row, int col)
    {
        int i;
    
        for(i=0; i<DIRECTSIZE; i++) {
            int nextrow = row + direct[i].drow;
            int nextcol = col + direct[i].dcol;
    
            if(grid[nextrow][nextcol] == '@') {
                grid[nextrow][nextcol] = '*';
    
                dfs(nextrow, nextcol);
            }
        }
    }
    
    int main(void)
    {
        int m, n, count, i, j;
    
        while(scanf("%d%d", &m, &n) != EOF) {
            // 判定结束条件
            if(m == 0 && n == 0)
                break;
    
            // 清零:边界清零
            memset(grid, 0, sizeof(grid));
    
            // 读入数据
            for(i=1; i<=m; i++)
                scanf("%s", grid[i]+1);
    
            // 计数清零
            count = 0;
    
            // 深度优先搜索
            for(i=1; i<=m; i++)
                for(j=1; j<=n; j++)
                    if(grid[i][j] == '@') {
                        count++;
                        grid[i][j] = '*';
                        dfs(i, j);
                    }
    
            // 输出结果
            printf("%d
    ", count);
        }
    
        return 0;
    }






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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563778.html
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