• ZOJ1002 Fire Net


    Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall.

    A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening.

    Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets.

    The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legalprovided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through.

    The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways.

    Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration.

     

    输入

     

    The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file. Each map description begins with a line containing a positive integer n that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a '.' indicating an open space and an uppercase 'X' indicating a wall. There are no spaces in the input file.

     

    输出

     

    For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.

     

    样例输入

    4
    .X..
    ....
    XX..
    ....
    2
    XX
    .X
    3
    .X.
    X.X
    .X.
    3
    ...
    .XX
    .XX
    4
    ....
    ....
    ....
    ....
    0

    样例输出

    5
    1
    5
    2
    4

    /*    错误的理解
    【我理解的题目意思】
    给出一个n*n(2<=n<=4)的矩阵,其中“.”表示路,“X”表示墙
    往里边放炸弹“o”【类似游戏炸弹人】,使得当炸弹一起爆炸的时候,矩阵都能够被炸到。
    但是炸弹不能够互相炸。也就是一颗炸弹的上下左右不能有炸弹,除非中间有墙隔着。
    【思路】
    类似八皇后问题,而且最多是4*4举证,就用回溯递归。
    第一批炸弹依次遍历第一排,如果有i堵墙的话,就放i+1个炸弹
     第二批炸弹遍历第二排
      ...
      第n批炸弹遍历最后排
    n批炸弹放完之后,就可以了~
    【题目只要求找出最多能放置多少个炸弹!我理解成了恰好摆满每个坐标!】
    */

    自己写了半天没有搞出来,最后看了“洞庭散人”的代码才明白了题意。。。

    http://www.cnblogs.com/phinecos/archive/2008/09/18/1293017.html

    ZOJ1002

    【重新理解题意】

    回溯过程:【递归(小规模问题)或循环(大规模问题)】

      1:能不能放

      2:放或者不放

    终止条件:

      没地方放

    其他处理:

      是不是最多。

    #include <stdio.h>
    #define true 1
    #define false 0
    char map[4][4];
    int maxNum,n;

    int CanPut(int row,int col)
    {
    int i;
    for (i=row-1;i>=0;i--)
    {
    if (map[i][col] == '0') return false;
    if (map[i][col] == 'X') break;
    }
    for (i=col-1; i>=0; i--)
    {
    if (map[row][i] == '0') return false;
    if (map[row][i] == 'X') break;
    }
    return true;
    }

    void Solve(int k,int curNum)
    {
    int x,y;
    if (k==n*n)
    {
    if (curNum>maxNum)
    {
    maxNum = curNum;
    return;
    }
    }
    else
    {
    x=k/n;
    y=k%n;
    if (map[x][y]=='.' && CanPut(x,y)==true)
    {
    map[x][y] = '0';
    Solve(k+1,curNum+1);
    map[x][y] = '.';
    }
    Solve(k+1,curNum);
    }
    }
    int main()
    {
    int i,j;
    while (scanf("%d",&n),n)
    {
    for (i=0;i<n;i++)
    {
    for (j=0;j<n;j++)
    {
    scanf("%d",&map[i][j]);
    }
    }
    maxNum = 0;
    Solve(0,0);
    printf("%d\n",maxNum);
    }
    return 0;
    }
    字节跳动内推

    找我内推: 字节跳动各种岗位
    作者: ZH奶酪(张贺)
    邮箱: cheesezh@qq.com
    出处: http://www.cnblogs.com/CheeseZH/
    * 本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。

  • 相关阅读:
    MySQL "show users"
    MySQL
    A MySQL 'create table' syntax example
    MySQL backup
    MySQL show status
    Tomcat, pathinfo, and servlets
    Servlet forward example
    Servlet redirect example
    Java servlet example
    How to forward from one JSP to another JSP
  • 原文地址:https://www.cnblogs.com/CheeseZH/p/2428557.html
Copyright © 2020-2023  润新知