Codeforces Round #315 (Div. 1) B. Symmetric and Transitive（Bell数的应用）（好题）
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 
, in this case we use a notation 
.
Binary relation is equivalence relation, if:
- It is reflexive (for any a it is true that
); - It is symmetric (for any a, b it is true that if , then
); - It is transitive (if and
, than 
).
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 , then (according to property (2)), which means (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.
A single line contains a single integer n (1 ≤ n ≤ 4000).
In a single line print the answer to the problem modulo 109 + 7.
1
1
2
3
3
10
If n = 1 there is only one such relation — an empty one, i.e. 
. In other words, for a single element x of set A the following is hold: 
.
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 ,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：
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 , in this case we use a notation .
Binary relation is equivalence relation, if:
- It is reflexive (for any a it is true that );（自反律）
- It is symmetric (for any a, b it is true that if , then ); （交换律）
- It is transitive (if and , than ).（传递律）
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 和 推出，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 second
typedef 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。
- 给定，有递归关系
证明递推式：考虑第n个物品，n可以单独构成一个非空集合，此时前n-1个物品构成k-1个非空的不可辨别的集合，有种方法；也可以前n-1种物品构成k个非空的不可辨别的集合，第n个物品放入任意一个中，这样有种方法。
Bell数：n的集合的划分方法的数目。
由定义可以得出它和第二类stirling数的关系：,即划分成k个非空集合
有两个递推关系可以求bell数：
1.
上述组合公式的证明：
可以这样来想，B_{n+1}是含有n+1个元素集合的划分的个数，考虑元素
假设他被单独划分到一类，那么还剩下n个元素，这种情况下划分个数为；
假设他和某一个元素被划分为一类，那么还剩下n-1个元素，这种情况下划分个数为 ；
假设他和某两个元素被划分为一类，那么还剩下n-2个元素，这种情况下划分个数为 ；
依次类推，得到了上述组合公式
2.根据贝尔三角形递推（较方便形象）
用以下方法建构一个三角矩阵（形式类似杨辉三角形）：
- 第一行第一项是1（）
- 对于n>1，第n行第一项等同第n-1行最后一项。（）
- 对于m,n>1，第n行第m项等于它左边和左上方的两个数之和。（）
结果如下：（OEIS:A011971）
每行首项是贝尔数。
摘自：
https://zh.wikipedia.org/wiki/%E8%B4%9D%E5%B0%94%E6%95%B0
https://zh.wikipedia.org/wiki/%E6%96%AF%E7%89%B9%E7%81%B5%E6%95%B0