[Codeforces626F] Group Projects (DP)

Group Projects

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Description

There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.

If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?

Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.

The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.

Output

Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo (10^9 + 7).

Sample Input

3 2
2 4 5

Sample Output

3

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题目大意

有n个商品，每个商品有不同的价值。要求把这些商品分组，每组有一个值为组内商品的最大价值差，问是这些每组值的和不超过m的方案数，答案对(1e9+7)取模。

这道题的定义比较难，首先我们我们对商品按其价值由小到大进行排序。我们可以发现，每一组的价值差为相邻商品价值差的和。
定义dp[i][j][k]表示前i件商品还有j组未完成，差值为k的分组种数。
共有4种转移方案

temp = (a[i] - a[i-1]) * j;

1. 第i件商品加入一个新分组，并且该分组未完成； dp[i][j+1][k] = dp[i][j+1][k] + dp[i-1][j][k-temp];
2. 第i件商品加入一个新分组，并且该分组只有一个元素；dp[i][j][k] = dp[i][j][k] + dp[i-1][j][k-temp];
3. 第i件商品加入一个之前的分组，并且该分组未完成； dp[i][j][k] = dp[i][j][k] + dp[i-1][j][k-temp] * j;
4. 第i件商品加入一个之前的分组，并且该分组已完成。 dp[i][j-1][k] = dp[i][j-1][k] + dp[i-1][j][k-temp] * j;

每一个转移都只与i 和 i-1有关，所以我们可以用滚动数组进行优化，
时间复杂度为(O(n^2k)), 空间复杂度为(nk).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int mod = 1e9 + 7;
int n,m;
int dp[2][210][1010];
int a[210];

int main(){
	ios::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m;
	for(int i = 1;i <= n;i++)cin >> a[i];
	sort(a + 1,a + n + 1);
	
	dp[0][0][0] = 1;
	int cur = 0;
	for(int i = 1;i <= n;i++){
		cur ^= 1;
		memset(dp[cur],0,sizeof(dp[cur]));
		int v = a[i] - a[i-1];
		for(int j = 0;j < i;j++){
			int temp = v * j;
			for(int k = temp;k <= m;k++){
				dp[cur][j + 1][k] = (dp[cur][j + 1][k] + dp[cur^1][j][k-temp]) % mod;
				dp[cur][j][k] = (dp[cur][j][k] + dp[cur^1][j][k-temp]) % mod;
				if(j)dp[cur][j-1][k] = (dp[cur][j-1][k] + (long long)dp[cur^1][j][k-temp] * j) % mod;
				if(j)dp[cur][j][k] = (dp[cur][j][k] + (long long)dp[cur^1][j][k-temp] * j) % mod;
			}
		}
	}
	int ans = 0;
	for(int i = 0;i <= m;i++)
		ans = (ans + dp[cur][0][i]) % mod;
	cout << ans << endl;
	return 0;
}