hdu5355 Cake

Problem Description

There are m soda and today is their birthday. The 1-st soda has prepared n cakes with size 1,2,…,n. Now 1-st soda wants to divide the cakes into m parts so that the total size of each part is equal. 

Note that you cannot divide a whole cake into small pieces that is each cake must be complete in the m parts. Each cake must belong to exact one of m parts.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (1≤n≤105,2≤m≤10), the number of cakes and the number of soda.
It is guaranteed that the total number of soda in the input doesn’t exceed 1000000. The number of test cases in the input doesn’t exceed 1000.

 

Output

For each test case, output "YES" (without the quotes) if it is possible, otherwise output "NO" in the first line.

If it is possible, then output m lines denoting the m parts. The first number si of i-th line is the number of cakes in i-th part. Then si numbers follow denoting the size of cakes in i-th part. If there are multiple solutions, print any of them.

 

Sample Input

4
1 2
5 3
5 2
9 3

 

Sample Output

NO
YES
1 5
2 1 4
2 2 3
NO
YES
3 1 5 9
3 2 6 7
3 3 4 8
这题和木棒拼接正方形非常像，用同样的思路即可了。
这里注意dfs可能比較深，所以要手动开栈。#pragma comment(linker, "/STACK:102400000,102400000") 这句话加在程序最前面。
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100050
#define ll long long
int vis[maxn],liang,fen,n;
set<int>myset[20];
set<int>::iterator it;

int dfs(int x,int pos,ll len)
{
	int i;
	if(x==fen)return 1;
	for(i=pos;i>=1;i--){
		if(!vis[i]){
			vis[i]=1;
			if(len+i<liang){
				myset[x].insert(i);
				if(dfs(x,i-1,len+i))return 1;
				myset[x].erase(i);
			}
			else if(len+i==liang){
				myset[x].insert(i);
				if(dfs(x+1,n,0))return 1;
				myset[x].insert(i);
			}
			vis[i]=0;
		}
	}
	return 0;
}


int main()
{
	int i,j,T;
	ll num;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&fen);
		num=(ll)(n+1)*n/2;
		if(n<fen || num%fen!=0 || num/fen<n){
			printf("NO
");continue;
		}
		liang=num/fen;
		memset(vis,0,sizeof(vis));
		for(i=0;i<=fen;i++){
			myset[i].clear();
		}
		
		if(dfs(0,n,0)){
			printf("YES
");
			for(i=0;i<fen;i++){
				printf("%d",myset[i].size());
				for(it=myset[i].begin();it!=myset[i].end();it++){
					printf(" %d",*it);
				}
				printf("
");
			}
		}
		else printf("NO
");
	}
	return 0;
}
/*
100
50 10
NO
40 10
YES
3 3 39 40
3 7 37 38
3 11 35 36
3 15 33 34
3 19 31 32
3 23 29 30
4 1 26 27 28
5 2 9 22 24 25
5 6 17 18 20 21
8 4 5 8 10 12 13 14 16
*/