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  • hdu 1045 Fire Net(dfs)

    Fire Net

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8239    Accepted Submission(s): 4734


    Problem Description
    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 legal provided 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.
     
    Input
    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.
     
    Output
    For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.
     
    Sample Input
    4
    .X..
    ....
    XX..
    ....
    2
    XX
    .X
    3
    .X.
    X.X
    .X.
    3
    ...
    .XX
    .XX
    4
    ....
    ....
    ....
    ....
    0
     
    Sample Output
    5
    1
    5
    2
    4
     
    Source
     
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    一道搜索题,因为数据很小,最大为4,所以可以直接深搜枚举,还是很不习惯写深搜,想了很久,最后参照别人的代码写的。
     
    题意:与八皇后类似,炮台不能存在于同行同列,除非中间隔墙。
     
     1 #include <iostream>
     2 #include <cstring>
     3 #include <cstdio>
     4 using namespace std;
     5 int n,s;
     6 char map[105][105];
     7 int can(int x,int y)
     8 {
     9     int i;
    10     for(i=x-1; i>=0; i--)
    11     {
    12         if(map[i][y]=='X')  //若出现墙,则可以直接跳出
    13             break;
    14         else if(map[i][y]=='K')
    15             return 0;
    16     }
    17     for(i=y-1; i>=0; i--)
    18     {
    19         if(map[x][i]=='X')
    20             break;
    21         else if(map[x][i]=='K')
    22             return 0;
    23     }
    24     return 1;
    25 }
    26 void DFS(int y,int count)
    27 {
    28     int i,j;
    29     if(count>s) //某一种方法的最大炮台数赋给s保存
    30         s=count;
    31     for(i=y; i<n; i++)   //每一行都要搜索
    32         for(j=0; j<n; j++)
    33         {
    34             if(map[i][j]=='.'&&can(i,j)) //can函数判断这个炮台是否能存在
    35             {
    36                 map[i][j]='K';  //放了炮台的标记为K
    37                 count++;
    38                 DFS(i,count); //进入下一次循环
    39                 count=0;
    40                 map[i][j]='.';
    41             }
    42         }
    43 }
    44 int main()
    45 {
    46     while(~scanf("%d",&n)&&n)
    47     {
    48         for(int i=0; i<n; i++)
    49             scanf("%s",map[i]);   //输入地图
    50         s=0;
    51         DFS(0,0);     //第0行开始搜,炮台数量s初始化为0
    52         printf("%d
    ",s);
    53     }
    54     return 0;
    55 }
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  • 原文地址:https://www.cnblogs.com/pshw/p/4813534.html
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