zoukankan      html  css  js  c++  java
  • 【杭电】[1241]Oil Deposits

    Oil Deposits

    Time Limit: 2000/1000 MS (Java/Others)

    Memory Limit: 65536/32768 K (Java/Others)

    Problem Description

    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.

    Input

    The input file 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.

    Output

    For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

    Sample Input

    1 1
    *
    3 5
    *@*@*
    **@**
    *@*@*
    1 8
    @@****@*
    5 5 
    ****@
    *@@*@
    *@**@
    @@@*@
    @@**@
    0 0

    Sample Output

    0
    1
    2
    2
    

    问几块石油区域
    需要注意的是只要八个方向上有石油
    那么这片区域就视作连通的
    所以dfs进行搜索填充就好

    #include<stdio.h>
    int move[2][8]= {{1,1,1,0,0,-1,-1,-1},{0,1,-1,1,-1,0,1,-1}};
    char map[120][120];
    int H,W;
    void dfs(int n,int m) {
        map[n][m]='*';
        for(int i=0; i<8; i++) {
            int tn=n+move[0][i],tm=m+move[1][i];
            if(tn>=0&&tn<H&&tm>=0&&tm<W&&map[tn][tm]=='@')
                dfs(tn,tm);
        }
    }
    int main() {
        while(scanf("%d %d",&H,&W),H||W) {
            for(int i=0; i<H; i++)
                scanf("%s",map[i]);
            int cnt=0;
            for(int i=0; i<H; i++) {
                for(int j=0; j<W; j++)
                    if(map[i][j]=='@') {
                        cnt++;
                        dfs(i,j);
                    }
            }
            printf("%d
    ",cnt);
        }
        return 0;
    }
    

    题目地址:【杭电】[1241]Oil Deposits


    查看原文:http://www.boiltask.com/blog/?p=1921
  • 相关阅读:
    关于setTimeout的妙用
    JavaScript中四种不同的属性检测方式比较
    AngularJS中transclude用法详解
    Token:服务端身份验证的流行方案
    浅析网页meta标签中X-UA-Compatible属性的使用
    谈谈近期学习Nativejs和reactNative的一些感受
    关于EasyUI DataGrid行编辑时嵌入时间控件
    全局程序集缓存工具(Gacutil.exe)用法详解
    JAVA从基础到框架搭建网站
    Swagger UI使用指南
  • 原文地址:https://www.cnblogs.com/BoilTask/p/12569456.html
Copyright © 2011-2022 走看看