2015暑假多校联合---Cake（深搜）

题目链接：HDU 5355

http://acm.split.hdu.edu.cn/showproblem.php?pid=5355

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

 

Author

zimpha@zju

 

Source

2015 Multi-University Training Contest 6

 

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题意：输入n，m  表示有n个蛋糕，大小为1~n，要均分给m个人，但是每一个蛋糕不能切分，给每个人的蛋糕必须是完整的，输出分配方案；

思路：刚开始，我是用贪心划分的，每个人应该划分的量h=（n+1）*n/2/m  首先给一个人分一个最大的蛋糕x，接下来分h-x，如果h-x已经给别人了，那么继续向下找h-x-1....

这样做是有bug的如23  6  这个数据就会输出NO  ，但实际上是有解的YES

3  11  12  23

4  1    10  13   22

4  2    9    14   21

4  3    8    15   20

4  4    7    16   19

4  5    6    17   18

故这样贪心的做法有bug；

正解：深搜枚举出所有的情况，找到正确的划分方案，这里有一个很好的可以优化，可以发现任取2*m个连续的数都可以均分给m个人，并且n>=2*m-1时才有划分方案，所以令nn=（n-2*m+1）%2*m+2*m-1  从nn+1到n是x*2*m，这部分很好划分，那么只需要对1~nn进行深搜即可；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <set>
using namespace std;
int n,m,nn,h;
bool vis[];

set<int>s[];
set<int>::iterator it;

int dfs(int cnt,int pos,int sum)
{
    if(cnt==m+) return ;
    for(int i=pos;i>=;i--)
    {
        if(!vis[i])
        {
            vis[i]=true;
            if(i+sum<h)
            {
               s[cnt].insert(i);
               if(dfs(cnt,pos-,sum+i)) return true;
               s[cnt].erase(i);
            }
            else if(i+sum==h)
            {
                s[cnt].insert(i);
                if(dfs(cnt+,n,)) return true;
                s[cnt].erase(i);
            }
            vis[i]=false;
        }
    }
    return ;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        for(int i=;i<;i++)
        s[i].clear();
        scanf("%d%d",&nn,&m);
        if(nn<*m-||(long long)(nn+)*nn/%m!=)   {puts("NO"); continue;}
        n=(nn-*m+)%(*m)+*m-;
        h=(n+)*n//m;
        memset(vis,false,sizeof(vis));

        dfs(,n,);

        int go=(nn-*m+)%(*m)+*m-;
        int  k=(nn-*m+)/(*m);
        for(int i=;i<=k;i++)
        {
            for(int j=;j<=m;j++)
            {
                s[j].insert(go+j);
                s[j].insert(go+*m+-j);
            }
            go+=*m;
        }
        puts("YES");
        for(int i=;i<=m;i++)
        {
            printf("%d",s[i].size());
            for(it=s[i].begin();it!=s[i].end();it++)
                printf(" %d",*it);
            puts("");
        }
    }
}