HDU - 1045 Fire Net

本文探讨了一个关于在包含墙壁的城市地图上放置防御塔的问题。目标是在合法配置下放置尽可能多的防御塔,使得任意两个防御塔不会互相射击到对方。文章提供了一种递归深度优先搜索的解决方案,并附带了完整的C++实现代码。

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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

题解:建房子问题,点(.)表示此地可以建房子,一般情况下,同行同列只能建一个房子,但是如果两房之间有墙相隔,则两个都可以存在。’X‘是墙,求最多能建多少个房子。

贴代码:

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
int n,ans;
char mp[5][5];
int check(int x,int y)
{
    for(int i=x-1; i>=0; i--)//检测此点上方
    {
        if(mp[i][y]=='X')
            break;
        if(mp[i][y]=='O')
            return 0;
    }
    for(int i=y-1; i>=0; i--)//左边
    {
        if(mp[x][i]=='X')
            break;
        if(mp[x][i]=='O')
            return 0;
    }
    return 1;
}
void dfs(int num,int sum)
{
    int x=num/n,y=num%n;第num个点的坐标
    if(num==n*n)
    {
        if(ans<sum)
            ans=sum;
        return ;
    }
    if(mp[x][y]=='.')
    {
        if(check(x,y))
        {
            mp[x][y]='O';//在此地建房子
            dfs(num+1,sum+1);
            mp[x][y]='.';
        }
    }
    dfs(num+1,sum);
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        int i,j;
        for(i=0; i<n; i++)
            scanf("%s",mp[i]);
        ans=-1;
        dfs(0,0);
        printf("%d\n",ans);
    }
    return 0;
}

 

### 关于HDU - 6609 的题目解析 由于当前未提供具体关于 HDU - 6609 题目的详细描述,以下是基于一般算法竞赛题型可能涉及的内容进行推测和解答。 #### 可能的题目背景 假设该题目属于动态规划类问题(类似于多重背包问题),其核心在于优化资源分配或路径选择。此类问题通常会给出一组物品及其属性(如重量、价值等)以及约束条件(如容量限制)。目标是最优地选取某些物品使得满足特定的目标函数[^2]。 #### 动态转移方程设计 如果此题确实是一个变种的背包问题,则可以采用如下状态定义方法: 设 `dp[i][j]` 表示前 i 种物品,在某种条件下达到 j 值时的最大收益或者最小代价。对于每一种新加入考虑范围内的物体 k ,更新规则可能是这样的形式: ```python for i in range(n): for s in range(V, w[k]-1, -1): dp[s] = max(dp[s], dp[s-w[k]] + v[k]) ``` 这里需要注意边界情况处理以及初始化设置合理值来保证计算准确性。 另外还有一种可能性就是它涉及到组合数学方面知识或者是图论最短路等相关知识点。如果是后者的话那么就需要构建相应的邻接表表示图形结构并通过Dijkstra/Bellman-Ford/Floyd-Warshall等经典算法求解两点间距离等问题了[^4]。 最后按照输出格式要求打印结果字符串"Case #X: Y"[^3]。 #### 示例代码片段 下面展示了一个简单的伪代码框架用于解决上述提到类型的DP问题: ```python def solve(): t=int(input()) res=[] cas=1 while(t>0): n,k=list(map(int,input().split())) # Initialize your data structures here ans=find_min_unhappiness() # Implement function find_min_unhappiness() res.append(f'Case #{cas}: {round(ans)}') cas+=1 t-=1 print("\n".join(res)) solve() ```
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