Babs’ Box Boutique(DFS)

本文探讨了如何通过深度优先搜索算法解决长方体堆叠问题,即在遵循尺寸限制条件下,找出能够堆叠的最大数量的长方体。文章提供了一个具体的实现案例,并通过示例输入输出展示了算法的有效性。

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题目链接:http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1943

题目大意:从小到大排列长方体,使上面的边界小于等于下面的;

题目思路:DFS全部寻找一遍

题目:

Babs’ Box Boutique

Time Limit: 1000MS Memory limit: 65536K

题目描述

Babs sells boxes and lots of them. All her boxes are rectangular but come in many different sizes. Babs wants to create a really eye-catching display by stacking, one on top of another, as many boxes as she can outside her store. To maintain neatness and stability, she will always have the sides of the boxes parallel but will never put a box on top of another if the top box sticks out over the bottom one. For example, a box with base 5-by-10 can not be placed on a box with base 12-by-4.
Of course the boxes have three dimensions and Babs can orient the boxes anyway she wishes. Thus a
5-by-10-by-12 box may be stacked so the base is 5-by-10, 5-by-12, or 10-by-12.
For example, if Babs currently has 4 boxes of dimensions 2-2-9, 6-5-5, 1-4-9, and 3-1-1, she could stack up to 3 boxes but not all four. (For example, the third box, the first box, then the last box, appropriately oriented. Alternatively, the second box could replace the third (bottom) box.)
Babs’ stock rotates, so the boxes she stacks outside change frequently. It’s just too much for Babs to figure out and so she has come to you for help. Your job is to find the most boxes Babs can stack up
given her current inventory. Babs will have no more than 10 different sized boxes and will use at most one box of any size in her display.

输入

A positive integer n (n ≤ 10) will be on the first input line for each test case. Each of the next n lines will contain three positive integers giving the dimensions of a box. No two boxes will have identical dimensions. None of the dimensions will exceed 20. A line with 0 will follow the last test case.
 

输出

For each test case, output the maximum number of boxes Babs can stack using the format given below.
 

示例输入

4
2 2 9
6 5 5
1 4 9
3 1 1
3
2 4 2
1 5 2
3 4 1
0

示例输出

Case 1: 3
Case 2: 3
代码:

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
struct node
{
    int x,y,z;
} w[15];
int n;
int visit[15];
int DFS(int a,int b,int num)
{
    int x1=num,x2=num,x3=num;
    for(int i=0; i<n; i++)
    {
        if(visit[i]==0)
        {
            if((w[i].x<=a&&w[i].y<=b)||(w[i].x<=b&&w[i].y<=a))
            {
                visit[i]=1;
                x1=max(x1,DFS(w[i].x,w[i].y,num+1));
            }
            visit[i]=0;
            if((w[i].x<=a&&w[i].z<=b)||(w[i].x<=b&&w[i].z<=a))
            {
                visit[i]=1;
                x2=max(x2,DFS(w[i].x,w[i].z,num+1));
            }
            visit[i]=0;
            if((w[i].y<=a&&w[i].z<=b)||(w[i].y<=b&&w[i].z<=a))
             {
                 visit[i]=1;
                 x3=max(x3,DFS(w[i].y,w[i].z,num+1));
             }
             visit[i]=0;
        }
    }
    int t=max(x1,x2);
    t=max(t,x3);
    num=t;
    //printf("%d %d %d %d\n",x1,x2,x3,num);
    return num;
}
int main()
{
    int fig=0;
    while(scanf("%d",&n))
    {
        if(n==0)
            break;
            fig++;
        memset(visit,0,sizeof(visit));
        for(int i=0; i<n; i++)
        {
            int aa,bb,cc;
            scanf("%d%d%d",&aa,&bb,&cc);
            w[i].x=aa;
            w[i].y=bb;
            w[i].z=cc;
        }
        printf("Case %d: %d\n",fig,DFS(25,25,0));
    }
    return 0;
}


资源下载链接为: https://pan.quark.cn/s/d9ef5828b597 在本文中,我们将探讨如何通过 Vue.js 实现一个带有动画效果的“回到顶部”功能。Vue.js 是一款用于构建用户界面的流行 JavaScript 框架,其组件化和响应式设计让实现这种交互功能变得十分便捷。 首先,我们来分析 HTML 代码。在这个示例中,存在一个 ID 为 back-to-top 的 div 元素,其中包含两个 span 标签,分别显示“回到”和“顶部”文字。该 div 元素绑定了 Vue.js 的 @click 事件处理器 backToTop,用于处理点击事件,同时还绑定了 v-show 指令来控制按钮的显示与隐藏。v-cloak 指令的作用是在 Vue 实例渲染完成之前隐藏该元素,避免出现闪烁现象。 CSS 部分(backTop.css)主要负责样式设计。它首先清除了一些默认的边距和填充,对 html 和 body 进行了全屏布局,并设置了相对定位。.back-to-top 类则定义了“回到顶部”按钮的样式,包括其位置、圆角、阴影、填充以及悬停时背景颜色的变化。此外,与 v-cloak 相关的 CSS 确保在 Vue 实例加载过程中隐藏该元素。每个 .page 类代表一个页面,每个页面的高度设置为 400px,用于模拟多页面的滚动效果。 接下来是 JavaScript 部分(backTop.js)。在这里,我们创建了一个 Vue 实例。实例的 el 属性指定 Vue 将挂载到的 DOM 元素(#back-to-top)。data 对象中包含三个属性:backTopShow 用于控制按钮的显示状态;backTopAllow 用于防止用户快速连续点击;backSeconds 定义了回到顶部所需的时间;showPx 则规定了滚动多少像素后显示“回到顶部”按钮。 在 V
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