Codeforces Round #315 (Div. 1) B. Symmetric and Transitive（Bell数的应用）（好题）

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Little Johnny has recently learned about set theory. Now he is studying binary relations. You've probably heard the term "equivalence relation". These relations are very important in many areas of mathematics. For example, the equality of the two numbers is an equivalence relation.

A set ρ of pairs (a, b) of elements of some set A is called a binary relation on set A. For two elements a and b of the set A we say that they are in relation ρ, if pair $\text{[math]}$ , in this case we use a notation $\text{[math]}$ .

Binary relation is equivalence relation, if:

It is reflexive (for any a it is true that $\text{[math]}$ );
It is symmetric (for any a, b it is true that if $\text{[math]}$ , then $\text{[math]}$ );
It is transitive (if $\text{[math]}$ and $\text{[math]}$ , than $\text{[math]}$ ).

Little Johnny is not completely a fool and he noticed that the first condition is not necessary! Here is his "proof":

Take any two elements, a and b. If $\text{[math]}$ , then $\text{[math]}$ (according to property (2)), which means $\text{[math]}$ (according to property (3)).

It's very simple, isn't it? However, you noticed that Johnny's "proof" is wrong, and decided to show him a lot of examples that prove him wrong.

Here's your task: count the number of binary relations over a set of size n such that they are symmetric, transitive, but not an equivalence relations (i.e. they are not reflexive).

Since their number may be very large (not 0, according to Little Johnny), print the remainder of integer division of this number by109 + 7.

Input

A single line contains a single integer n (1 ≤ n ≤ 4000).

Output

In a single line print the answer to the problem modulo 109 + 7.

Sample test(s)
input
1
output
1
input
2
output
3
input
3
output
10

Note

If n = 1 there is only one such relation — an empty one, i.e. $\text{[math]}$ . In other words, for a single element x of set A the following is hold: $\text{[math]}$ .

If n = 2 there are three such relations. Let's assume that set A consists of two elements, x and y. Then the valid relations are $\text{[math]}$ ,ρ = {(x, x)}, ρ = {(y, y)}. It is easy to see that the three listed binary relations are symmetric and transitive relations, but they are not equivalence relations.

大致题意：

Binary relation is equivalence relation, if:

It is reflexive (for any a it is true that $\text{[math]}$ );（自反律）
It is symmetric (for any a, b it is true that if $\text{[math]}$ , then $\text{[math]}$ ); （交换律）
It is transitive (if $\text{[math]}$ and $\text{[math]}$ , than $\text{[math]}$ ).（传递律）

翻译一下就是：数的全集为A，一些A中的元素组成的二元组是ρ 集合的元素，对于ρ 中任意元素pair(a,b)来说，他们满足关系 $\text{[math]}$

给出了等值Binary relation的定义，给定数的全集｛1，2，3...,n｝求有多少个ρ 集合满足等值关系中的2，3两条而不满足第一条

题意真是难理解...

思路：

只要有某个元素a满足了2，3两条，就一定由 $\text{[math]}$ 和 $\text{[math]}$ 推出 $\text{[math]}$ ，所以这个元素就满足了第一条

同理如果这个集合中的每个元素满足2，3条那么每个元素都满足第一条也就是此集合成为了等值Binary relation。

所以为了不成为等值Binary relation，必须至少有一个元素x不满足所有关系，也就是不参与构成了某个二元组，除x外的元素恰好构成了等值Binary relation。

所以本题的答案就是 sigama(i:0,n-1)｛ C(n,i)*(这i元素构成的等值Binary relation的个数) ｝

现在问题变成i个元素可以构成多少个等值Binary relation

这个问题可以用递推解决，求dp[i]，对于第i个元素必须满足这三个条件才行，所以必须让第i个元素与i-1个元素中的k个元素“混合”才行

比如{(a~b),(b~a),(a~a)}混入c后要满足等值Binary relation则变成{(a~b),(b~a),(b~c),(c~b),(a~c),(c~a),(a~a),(b~b),(c~c)}只有唯一的混合结果

所以dp[i] = sigma(k:0,i-1){ C(i-1,k)*dp[i-1-k] } = sigma(k:0,i-1){ C(i-1,i-1-k)*dp[i-1-k] } = sigma(k:0,i-1){ C(i-1,k)*dp[k] }

会发现这就是Bell数的递推公式

所以求“i个元素可以构成多少个等值Binary relation”这个问题等价与“i的集合的划分方法的数目",例如：

集合{a, b, c}有5种不同的划分方法：

{{a}, {b}, {c}}

{{a}, {b, c}}

{{b}, {a, c}}

{{c}, {a, b}}

{{a, b, c}};

每个块中都是前面所说的混合构造成等值Binary relation，所以知道bell数的会很容易解决此问题，到此问题已经圆满解决

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <ctime>#include <bitset>#include <algorithm>#define SZ(x) ((int)(x).size())#define ALL(v) (v).begin(), (v).end()#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i)#define REP(i,n) for ( int i=1; i<=int(n); i++ )#define rep(i,n) for ( int i=0; i< int(n); i++ )using namespace std;typedef long long ll;#define X first#define Y secondtypedef pair<int,int> pii;template <class T>inline bool RD(T &ret) {    char c; int sgn;    if (c = getchar(), c == EOF) return 0;    while (c != '-' && (c<'0' || c>'9')) c = getchar();    sgn = (c == '-') ? -1 : 1;    ret = (c == '-') ? 0 : (c - '0');    while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');    ret *= sgn;    return 1;}template <class T>inline void PT(T x) {    if (x < 0) {        putchar('-');        x = -x;    }    if (x > 9) PT(x / 10);    putchar(x % 10 + '0');}const int N = 4000+123;const int MOD = 1e9+7;int bell[N][N];int C[N][N];int main(){        bell[0][0] = 1;        REP(i,N-10)                REP(j,i){                        if(j == 1) bell[i][j] = bell[i-1][i-1];                        else bell[i][j] = (bell[i][j-1]+bell[i-1][j-1])%MOD;                }        REP(i,N-10) {                C[i][i] = C[i][0] = 1;                REP(j,i-1) C[i][j] = (C[i-1][j-1]+C[i-1][j])%MOD;        }        int n;        while(~scanf("%d",&n)){                ll ans = 0;                rep(i,n) ans = (ans+C[n][i]*(ll)bell[i][i])%MOD;                PT(ans),puts("");        }}

作为弱渣第一次遇过bell数，现在补充记录一些常用的基础知识：

Stirling数，第一类有点难理解不在此记录

第二类Stirling数是 $n$ 个元素的划分成k个非空集合的方法数目。常用的表示方法有 $S(n,k) , S_n^{(k)} , \left\{\begin{matrix} n \\ k \end{matrix}\right\}$ 。