【DFS】Anniversary Cake

[poj1020]Anniversary Cake

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17203		Accepted: 5619

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

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

Sample Output

KHOOOOB!
HUTUTU!

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

 

题目大意：给出一个蛋糕的长，是否能切成M个给定的小正方形且不留剩余？

 

试题分析：很显然，如果所有小蛋糕的面积大于大蛋糕的面积，那么就不行

                                  剩下的枚举一下高度，尽量先放大的，往下放，维护每一列的高度就可以了

 

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
//#include<cmath>

using namespace std;
const int INF = 9999999;
#define LL long long

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
int N,M;
int T;
int a[60];
int sum;
int maxr[60];
int re[60];

bool DFS(int st){
	if(st==M){return true;}
	int Minn=INF,tmp;
	for(int i=1;i<=N;i++){
		if(maxr[i]<Minn){
			tmp=i;
			Minn=maxr[i];
		}
	}
	for(int i=10;i>=1;i--){
		if(!re[i]) continue;
		bool canin;
		if(N-maxr[tmp]>=i&&N-tmp+1>=i){
			canin=true;
			for(int j=tmp;j<tmp+i;j++){
				if(maxr[j]>maxr[tmp]) {
				    canin=false;
				    break;
				}
			}
			if(canin){
				for(int j=tmp;j<tmp+i;j++) maxr[j]+=i;
				re[i]--;
				if(DFS(st+1)) return true;
				re[i]++;
				for(int j=tmp;j<tmp+i;j++) maxr[j]-=i;
			}
		}
	}
	return false;
}

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	T=read();
	while(T--){
		memset(re,0,sizeof(re));
		memset(maxr,0,sizeof(maxr));
		memset(a,0,sizeof(a));
		sum=0;
		N=read(),M=read();
		for(int i=1;i<=M;i++) a[i]=read(),sum+=(a[i]*a[i]),re[a[i]]++;
		if(sum!=N*N) {printf("HUTUTU!\n");continue;}
		if(DFS(0)) printf("KHOOOOB!\n");
		else printf("HUTUTU!\n");
	}
	return 0;
}