1737

      楼教主推荐的“男人八题”的第一题，要用到组合数学+大数运算。

 

      大意就是给你n个点，可以组成多少个连通图。解题思路是这样：设n个点的集合V可以组成G(n)个连通图，V分割成分别有m个点和n－m个点的两个子集V1、V2，如果两个子集内部都组成了连通图，那么V1、V2之间只要有若干个点连在一起就是一个大的连通图。为了避免重复，V1、V2各选定一点分别为p1、p2，V1至少一点连向p2，总共有2^m-1种情况。由于p1、p2都是选定的，所以V1、V2的分割的情况有C(m-1,n-2)种。所以这样最后的公式就是G(n)=sum(G(m)*G(m-n)*C(m-1,n-2)*(2^m-1))，1<=m<=n-1。

 

      排列组合的题目找出一个不重复的方案有时真的不容易。如果V1、V2之间任意两点都可以连，就会有重复。因为假设V分成3个子集V1'、V2'、V3'，第一次V1'、V2'分在V1并组成连通图，V3'在V2，然后V1、V2之间任一点互联；第二次V1'、V3'分在V1，V2'在V2，然后。。。。就重复了。所以最彻底的做法，指定两个点p1、p2分别放在V1、V2，其他点选择“依附”于其中之一。然后V1所有点选择与p2连或不连(不过至少连一个)。不同子集的点只能通过p2相连，或者被分在同一子集时就可以直接连了。这避免了重复。

 

       因为里面的数字都是天文数字，所以要用到大数运算，基本上等于把1002重做一遍。不过任意位整数的加减乘除做成一个库也蛮好。

 

 

#include <iostream>#include <algorithm>#include <vector>#include <string>#include <iterator>using namespace std;class add_carry{public:add_carry(): carry(0){};void operator()(unsigned char &x){x += carry;if(x > 9){x -= 10;carry = 1;}else{carry = 0;}}private:unsigned char carry;};template<typename T>class multi_factor{public:multi_factor(T x): factor(x), carry(0){};void operator()(unsigned char &x){carry += x * factor;if(carry > 9){x = carry % 10;carry /= 10;}else{x = carry;carry = 0;}}private:const T factor;T carry;};template<typename T>class multi_carry{public:multi_carry(): carry(0){};void operator()(T &x){x += carry;if(x > 9){carry = x / 10;x %= 10;}else{carry = 0;}}private:T carry;};template<typename T>class predicate{public:bool operator()(T x){return x != 0;}};class large_num{public:template<typename T>large_num(T x);void operator += (const large_num &x);void assign(unsigned long long x);template<typename T>void operator *= (const T &x);friend ostream &operator << (ostream &oput, const large_num &right);void power(int x);private:vector<unsigned char> data;};template<typename T>large_num::large_num(T x){if(x){while(x){data.push_back(x % 10);x /= 10;}}else{data.push_back(0);}}void large_num::power(int x){data.clear();unsigned long long temp = 1;temp <<= x;--temp;while(temp){data.push_back(temp % 10);temp /= 10;}}void large_num::operator += (const large_num &x){data.size() >= x.data.size()? data.push_back(0): data.resize(1 + x.data.size(), 0);for(size_t i = 0; i < x.data.size(); ++i){data[i] += x.data[i];}for_each(data.begin(), data.end(), add_carry());if(!data.back()){data.pop_back();}}void large_num::assign(unsigned long long x){data.clear();while(x){data.push_back(x % 10);x /= 10;}}template<typename T>void large_num::operator *= (const T &x){data.resize(sizeof(T) * 3 + data.size(), 0);for_each(data.begin(), data.end(), multi_factor<T>(x));vector<unsigned char>::reverse_iterator rpos = find_if(data.rbegin(), data.rend(), predicate<unsigned char>());vector<unsigned char>::iterator pos = rpos.base();data.resize(pos - data.begin());}template<>void large_num::operator *= <large_num> (const large_num &x){vector<unsigned short> temp;vector<large_num> buf(10, 0);temp.resize(data.size() + x.data.size(), 0);for(size_t i = 0; i < x.data.size(); ++i){unsigned char index = x.data[i];if(index){if(1 == buf[index].data.size() && !buf[index].data[0]){buf[index].data = data;buf[index] *= x.data[i];}for(size_t j = 0; j < buf[index].data.size(); ++j){temp[j + i] += buf[index].data[j];}}}for_each(temp.begin(), temp.end(), multi_carry<unsigned short>());vector<unsigned short>::reverse_iterator rpos = find_if(temp.rbegin(), temp.rend(), predicate<unsigned short>());vector<unsigned short>::iterator pos = rpos.base();data.resize(pos - temp.begin());for(size_t i = 0; i < data.size(); ++i){data[i] = static_cast<unsigned char>(temp[i]);}}ostream &operator << (ostream &oput, const large_num &right){copy(right.data.rbegin(), right.data.rend(), ostream_iterator<int>(oput));return oput;}int main(){int input;vector<unsigned long long> c_last(51, 0);vector<unsigned long long> c_current(51, 0);c_current[0] = 1;vector<large_num> graph(51, 0);graph[1].assign(1);graph[2].assign(1);int last = 3;large_num temp(1);while(cin >> input, input){for(; last <= input; ++last){c_last.assign(c_current.begin(), c_current.begin() + last - 2);for(int i = 1; i < last - 2; ++i){c_current[i] = c_last[i - 1] + c_last[i];}c_current[last - 2] = 1;for(int m = 1; m <= last - 1; ++m){temp.power(m);temp *= c_current[m - 1];temp *= graph[m];temp *= graph[last - m];graph[last] += temp;}}cout << graph[input] << endl;}return 0;}