Cake

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3255 Accepted Submission(s): 689
Special Judge

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

(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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#include <algorithm>
#include <iostream>
#include<vector>
#include<cstring>
#include<stdio.h>
#define maxn 100005
using namespace std;
long long  n,m,tar,res,tim;
int vis[maxn],plan[maxn];
vector<int> ans[100];
/*
分配的构造技巧，
刚开始想用贪心的后来无疑被wrong,
然后这题输入和数据范围上都要预判一下，
最好直接用long  long拉倒。

 参考了网上的思路，
 对其进行分组构造，
 每2*m个构成一组，对每组进行取值，
 可以构造出m个相同的大小。
 
 最后的回溯区间应该是4*m，
 。。。暂且还不知道为什么。。。
*/
bool dfs(int pos,int ss,int mm)
{
    if(mm==m)   return  true;
    for(int i=res;i>=pos;i--)
    {
        if(vis[i]==1) continue;
        if(ss+i==tar)
        {
            plan[i]=mm;
            //ans[mm].push_back(i);
            vis[i]=1;
            if( dfs(1,0,mm+1) ) return true;
            vis[i]=0;
            //ans[mm].pop_back();
        }
        else if(ss+i<tar)
        {
            vis[i]=1;
            plan[i]=mm;
            //ans[mm].push_back(i);
            if(dfs(i,ss+i,mm)) return true;
            vis[i]=0;
            //ans[mm].pop_back();
        }
    }
    return false;
}

int main()
{
    ios::sync_with_stdio(false);
    int t;scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        memset(plan,0,sizeof(plan));
        scanf("%ld%ld",&n,&m);
        if(  (n*(n+1)/2) %m!=0 )
        {
            printf("NO\n");
           continue;
        }
           long long s=n*(n+1)/2;
           tar=s/m;
           if(tar<n)
           {
               puts("NO");
               //cout<<"NO"<<endl;
               continue;
           }
           res=n%(2*m);
           if(res!=0)
           {
               res=res+2*m;
               res=min(res,n);
           }
           for(int i=n;i>res;i-=2*m)
           {
               int rb=i,lb=rb-2*m+1,tmp=0;
               tar-=(lb+rb);
               while(lb<=rb)
               {
                   ans[tmp].push_back(lb);
                   ans[tmp++].push_back(rb);
                   lb++,rb--;
               }
            }
            puts("YES");
            dfs(1,0,0);
            for(int i=1;i<=res;i++)
                 ans[plan[i]].push_back(i);
            for(int i=0;i<m;i++)
            {
                long long  tmp=ans[i].size();
                printf("%d",tmp);
                for(int j=0;j<tmp;j++)
                    printf(" %d",ans[i][j]);
                puts("");
            }
       for(int i=0;i<=m;i++) ans[i].clear();
     }
    return 0;
}

HDU 5355 Cake (构造+简单回溯)

Cake

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