zoukankan      html  css  js  c++  java
  • ACM-

    解题思路:Find Gold

    Someone locked you in the grid house. You can’t escape, until you find the enough gold. But somewhere is on the fire, so you can’t stand on this place. You can go with eight direction: up down, left, right, left-up, left-down, right-up, right-down. Sure, your first position is not fired.

    输入

    The input will contain one or more grid. Each grid is preceded by a line containing the number of rows and columns in the grid and the row and column of the start position. All numbers are in the range 1-25. The rows of the grid follow, starting on the next line, consisting of 'O','.' and 'X' characters. 'O' stand for gold, '.' stand for fire, 'X' stand for there are nothing . The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks.

    输出

    For each grid in the input, the output contains a single line with the max number of the gold you can get.

    样例输入

    4 6 2 2
    . . . X X X 
    . X . O . O 
    X . . . . . 
    X X X . . . 
    6 6 4 5
    X X O . X O 
    O O X . O O 
    . . . . . . 
    X X X . O O 
    X . X . . . 
    O X X . . O 
    4 7 1 2
    . X . . . . X 
    . X X . X X . 
    . . X X O O X 
    . O X X . X O
    0 0 0 0
    

    样例输出

    0
    2
    4

    深搜的水题。。。。。唯一麻烦的字符串的读入读出,为了避免麻烦,最好还是用cincout比较好。

    // Find Gold.cpp : 定义控制台应用程序的入口点。
    //
    
    #include "stdafx.h"
    
    #include <stdio.h>
    #include <string.h>
    #include <string>
    #include <iostream>
    using namespace std;
    const int M = 10005;
    int n, m, sr, sl, ans, vis[M][M], dir[8][2] = { 1, 0, -1, 0, 0, 1, 0, -1, 1, 1, -1, -1, 1, -1, -1, 1 };
    char map[M][M];
    
    void DFS(int x, int y)
    {
        vis[x][y] = 1;
        for (int i = 0; i < 8; i++)
        {
            int nx = dir[i][0] + x;
            int ny = dir[i][1] + y;
            if (nx>=0 && ny>=0 && nx<n && ny<m && !vis[nx][ny] && map[nx][ny] != '.')
            {
                if (map[nx][ny] == 'O')
                    ans++;
                DFS(nx, ny);
            }
        }
    }
    
    int main()
    {
        while (scanf("%d %d %d %d", &n, &m, &sr, &sl))
        {
            if (n == 0 && m == 0 && sr == 0 && sl == 0) break;
            memset(vis, 0, sizeof(vis));
            memset(map, '', sizeof(map));
            ans = 0;
            for (int i = 0; i < n; i++)
            {
                getchar();//读取回车
                for (int j = 0; j < m; j++)
                {
                    char c = getchar();
                    map[i][j] = c;
                    getchar();//读取空格
                }
            }
            if (map[sr - 1][sl - 1] == 'O') ans++;
            DFS(sr - 1, sl - 1);
            printf("%d
    ", ans);
    
        }
        return 0;
    }
  • 相关阅读:
    git 记录
    js 技巧
    首页三张幻灯片放什么
    6系统盈利方向
    WordPress获取某个分类关联的标签
    目标型长尾如何优化
    简介如何去除WordPress主题版权保护的方法
    2017.7.7 长尾关键词系统优化
    2017.7.6 目标关键词分析
    2017.7.2 seo知识总纲
  • 原文地址:https://www.cnblogs.com/x739400043/p/8505465.html
Copyright © 2011-2022 走看看