poj1020 Anniversary Cake 搜索

Description

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

24 8 1 1 1 1 1 3 1 15 6 3 3 2 1 1 1

Sample Output

KHOOOOB!HUTUTU!题意：有一块面积M X M的平面蛋糕 问是否能分成N块给定面积为a X a的蛋糕首先应该换个思路理解题目   能否分成N块  ==  能否恰用N块填充满然后再按常规搜索思路给出填充策略 ：1.先下后上 先左后右 2.先填充大的蛋糕再填充小的接着开始搜索了， 直接将蛋糕理解为二维数组去搜索是不行的耗时而且麻烦我们可以给出一个数组 column[N] 表示每列填充了几个1X1的小蛋糕  要判断能否填充一个a X a蛋糕（假设从J列开始）  只需要判断col[j]~col[j+a-1]是否都满足 M-col[x]>=a 即可 按上述策略搜索即可得到答案另注意两个剪枝：大蛋糕的面积！=所有小蛋糕的面积之和时    肯定不行边长比大蛋糕的边长的二分之一长的小蛋糕数量大于一时肯定不行 （可画图理解） 代码：
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int sumsize,n;int col[45],sizenum[12]; //MAXarea=1600    sizenum 表示这个size的数量int dfs(int filled){    if(filled==n)        return 1;    int i,j,minn=20;    for(i=1; i<=sumsize; i++)   //找填充最少的列开始填充    {        if(col[i]<minn)        {            minn=col[i];            j=i;        }    }    for(int size=10;size>=1;size--)    {        if(sizenum[size])             {            if(sumsize-minn>=size&&sumsize+1-j>=size)            {                int wide=0;                for(i=j;i<j+size;i++)  //判断横方向能否填充                {                    if(col[i]<=minn)                    {                        wide++;                        continue;                    }                    break;                }                if(wide>=size)                {                    sizenum[size]--;        //填充                    for(i=j;i<j+size;i++)                        col[i]+=size;                    if(dfs(filled+1))                        return 1;                                            sizenum[size]++;         //回溯                    for(i=j;i<j+size;i++)                        col[i]-=size;                }            }        }    }    return 0;}int main(){    int t;    int size;    int area,s;    scanf("%d",&t);    while(t--)    {        s=area=0;        memset(sizenum,0,sizeof(sizenum));        memset(col,0,sizeof(col));        scanf("%d%d",&sumsize,&n);        for(int i=1; i<=n; i++)        {            scanf("%d",&size);            sizenum[size]++;            if(size>sumsize/2)                s++;            area+=size*size;        }        if(s>1||area!=sumsize*sumsize)       //两个剪枝        {            cout<<"HUTUTU!"<<endl;            continue;        }        else        {            if(dfs(0))                cout<<"KHOOOOB!"<<endl;            else                cout<<"HUTUTU!"<<endl;        }    }    return 0;}