HDU 5355 Cake(2015多校第六场，搜索 + 剪枝）

Cake

Time Limit: 2000/1000 MS (Java/Others)    Memory
Limit: 131072/131072 K (Java/Others)

Total Submission(s): 965    Accepted Submission(s): 119

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, \dots, 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
\le n \le 10^5, 2 \le m \le 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 s_i of i-th
line is the number of cakes in i-th
part. Then s_inumbers
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

 

Source

2015 Multi-University Training Contest 6

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <algorithm>
#define LL long long
using namespace std;
const LL MAXN = 500000 + 10;
LL n, m;
LL ans[20][MAXN];
LL A[MAXN];
LL pos[MAXN];
LL tot, tar;
bool dfs(LL dep, LL now, LL u, LL c)
{
    if(now == 0)
    {
        LL k = 0;
        while(pos[k] != -1) k++;
        pos[k] = c;
        if(dfs(dep + 1, A[k], k + 1, c)) return true;
        pos[k] = -1;
        return false;
    }
    if(now == tar)
    {
        if(dep == tot) return true;
        if(dfs(dep, 0, 0, c + 1)) return true;
    }
    for(LL i=u;i<tot;i++)
    {
        if(pos[i] == -1 && now + A[i] <= tar)
        {
            pos[i] = c;
            if(dfs(dep + 1, now + A[i], i + 1, c)) return true;
            pos[i] = -1;
        }
    }
    return  false;
}
int main()
{
    LL T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(LL i=0;i<m;i++) for(LL j=0;j<n;j++) ans[i][j] = 0;
        if((n * (n + 1) / 2) % m != 0 || (2 * m - 1) > n)
        {
            printf("NO\n");
            continue;
        }
        while(n >= 40)
        {
            for(LL i=0;i<m;i++) ans[i][++ans[i][0]] = n - i;
            for(LL i=0;i<m;i++) ans[i][++ans[i][0]] = n - 2 * m + 1 + i;
            n -= 2 * m;
        }
        tot = n;
        tar = n * (n + 1) / (2 * m);
        for(LL i=0;i<tot;i++) A[i] = tot - i;
        for(LL i=0;i<tot;i++) pos[i] = -1;
        dfs(0, 0, 0, 0);
        for(LL i=0;i<tot;i++) ans[pos[i]][++ans[pos[i]][0]] = A[i];
        printf("YES\n");
        for(LL i=0;i<m;i++)
        {
            printf("%d", ans[i][0]);
            for(LL j=1;j<=ans[i][0];j++) printf(" %d", ans[i][j]);
            printf("\n");
        }
    }
    return 0;
}