Anniversary (DFS)

最新推荐文章于 2021-01-17 11:04:43 发布

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最新推荐文章于 2021-01-17 11:04:43 发布

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分类专栏：搜索文章标签： DFS

本文链接：https://blog.youkuaiyun.com/ZCMU_2024/article/details/89388577

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本文介绍了一种解决正方形蛋糕是否能被切成特定大小小块的算法，通过深度优先搜索（DFS）策略，确保每个小块都是正方形且没有浪费。文章详细解释了算法思路，包括如何选择填充顺序和条件，以及如何判断解决方案的存在性。

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Nahid Khaleh decides to invite the kids of the "Shahr-e Ghashang" to her wedding anniversary. She wants to prepare a square-shaped chocolate cake with known size. She asks each invited person to determine the size of the piece of cake that he/she wants (which should also be square-shaped). She knows that Mr. Kavoosi would not bear any wasting of the cake. She wants to know whether she can make a square cake with that size that serves everybody exactly with the requested size, and without any waste.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by input data for each test case. Each test case consist of a single line containing an integer s, the side of the cake, followed by an integer n (1 <= n <= 16), the number of cake pieces, followed by n integers (in the range 1...10) specifying the side of each piece.

Output

There should be one output line per test case containing one of the words KHOOOOB! or HUTUTU! depending on whether the cake can be cut into pieces of specified size without any waste or not.

Sample Input

2
4 8 1 1 1 1 1 3 1 1
5 6 3 3 2 1 1 1

Sample Output

KHOOOOB!
HUTUTU!

题意:问n个小正方型能否组成一个面积为side*side的大正方

思路:见代码

/*思路: DFS
1.我选择按列来放,保证每次填充时选则填充的列剩余未填充的行数>=指定蛋糕的边长
2.假设我们的列的编号是从1开始的，那么我们就要保证当前选定列+选择填充的蛋糕边长-1<=蛋糕的边长
3.有两种情况是直接可以判断结果为"HUTUTU!"的
  1)所有小蛋糕面积的总和!=大蛋糕的面积
  2)存在一个以上小蛋糕的边长>大蛋糕边长/2
*/
#include<bits/stdc++.h>
using namespace std;
const int Max=1e3+5;
int pos[100],cake[20],side,n; //pos记录第i列的开销,cake记录各种边长蛋糕的个数
/*首先明确填充的顺序:
  1.从左往右填充
  2.从下到上填充
  3.选择蛋糕先大后小
*/
bool DFS(int number)
{
    if(number==n)
        return true;
    int Min=Max;
    int index;
    for(int i=1;i<=side;i+=1)
    {
        if(Min>pos[i])
        {
            Min=pos[i];
            index=i;
        }
    }
    for(int i=10;i>=1;i--)
    {
        /* if((如果不存在边长为i的蛋糕)||(选定列已填充的宽度+将要填充的宽度>给定蛋糕的尺寸)||
           (从当前列开始向右填充的长度>给定蛋糕的尺寸)) 跳过;
        */
        if(!cake[i]||Min+i>side||index+i-1>side)
            continue;
        bool sign=true;
        for(int x=index;x<=index+i-1;x+=1)
        {
            /*从选定列遍历到终止的列,若其中发现某列的开销大于选定列的开销则放弃此次填充
              (原因:因为实际上填充的是一个完整正方形)
            */
            if(pos[x]>Min)
            {
                sign=false;
                break;
            }
        }
        if(!sign)
            continue;
        cake[i]--;
        for(int x=index;x<=index+i-1;x+=1)
            pos[x]+=i;
        if(DFS(number+1))
            return true;
        cake[i]+=1;
        for(int x=index;x<=index+i-1;x+=1)
            pos[x]-=i;
    }
    return false;
}
int main()
{
    std::ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        int pie[20];
        cin>>side>>n;
        int total=0;
        int Count=0;
        memset(cake,0,sizeof(cake));
        memset(pos,0,sizeof(pos));
        for(int i=0;i<n;i+=1)
        {
            cin>>pie[i];
            cake[pie[i]]+=1;
            total+=pie[i]*pie[i];
            if(pie[i]>side/2)
                Count+=1;
        }
        if(total!=side*side||Count>1)
            cout<<"HUTUTU!"<<endl;
        else
        {
            if(DFS(0))
                cout<<"KHOOOOB!"<<endl;
            else
                cout<<"HUTUTU!"<<endl;
        }
    }
    return 0;
}