Connected Graph POJ1737 高精度

题目链接

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An undirected graph is a set V of vertices and a set of E∈{V*V} edges.An undirected graph is connected if and only if for every pair (u,v) of vertices,u is reachable from v.You are to write a program that tries to calculate the number of different connected undirected graph with n vertices.For example,there are 4 different connected undirected graphs with 3 vertices.

Input

The input contains several test cases. Each test case contains an integer n, denoting the number of vertices. You may assume that 1<=n<=50. The last test case is followed by one zero. 

Output

For each test case output the answer on a single line. 

Sample Input

12340

Sample Output

11438

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题意:

给你一个无向图,问你有多少种不同的方式使这个图连通.

思路:

n 最大是50, 所以这个题肯定会爆long long,所以需要高精度, 那么接下来就是如何计算答案了.

首先我们知道 ans[1] = ans[2] = 1;
n >= 3 时,我们考虑拿掉点 1 和点 2 的情况, 假设点 2 所在的连通块共 k 个点,这 k 个点与剩下的 n - k 个点分别处于一个连通块中, 共有ans[k] * ans[n-k] 种, 点 2 所在的连通块是需要从除去点 1 和点 2 剩下的点中选取 k-1 个点,则有C(n-2,k-1)种,而这 k 个点必须通过点 1 才能与点 1 所在的连通块相连,于是有2 ^k - 1 种,
所以答案为:
ans(n)=Sum(ans(k)∗ans(n−k)∗C(n−2,k−1)∗(2k−1)|1<=k<n)

代码:

#include <algorithm>#include <cassert>#include <cstdio>#include <cstring>#include <iostream>#include <string>#include <vector>#include <cmath>using namespace std;struct BigInteger {    typedef unsigned long long LL;    static const int BASE = 100000000;    static const int WIDTH = 8;    vector<int> s;    BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}    BigInteger(LL num = 0) {*this = num;}    BigInteger(string s) {*this = s;}    BigInteger& operator = (long long num) {        s.clear();        do {            s.push_back(num % BASE);            num /= BASE;        } while (num > 0);        return *this;    }    BigInteger& operator = (const string& str) {        s.clear();        int x, len = (str.length() - 1) / WIDTH + 1;        for (int i = 0; i < len; i++) {            int end = str.length() - i*WIDTH;            int start = max(0, end - WIDTH);            sscanf(str.substr(start,end-start).c_str(), "%d", &x);            s.push_back(x);        }        return (*this).clean();    }    BigInteger operator + (const BigInteger& b) const {        BigInteger c; c.s.clear();        for (int i = 0, g = 0; ; i++) {            if (g == 0 && i >= s.size() && i >= b.s.size()) break;            int x = g;            if (i < s.size()) x += s[i];            if (i < b.s.size()) x += b.s[i];            c.s.push_back(x % BASE);            g = x / BASE;        }        return c;    }    BigInteger operator - (const BigInteger& b) const {        assert(b <= *this); // 减数不能大于被减数        BigInteger c; c.s.clear();        for (int i = 0, g = 0; ; i++) {            if (g == 0 && i >= s.size() && i >= b.s.size()) break;            int x = s[i] + g;            if (i < b.s.size()) x -= b.s[i];            if (x < 0) {g = -1; x += BASE;} else g = 0;            c.s.push_back(x);        }        return c.clean();    }    BigInteger operator * (const BigInteger& b) const {        int i, j; LL g;        vector<LL> v(s.size()+b.s.size(), 0);        BigInteger c; c.s.clear();        for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];        for (i = 0, g = 0; ; i++) {            if (g ==0 && i >= v.size()) break;            LL x = v[i] + g;            c.s.push_back(x % BASE);            g = x / BASE;        }        return c.clean();    }    BigInteger operator / (const BigInteger& b) const {        assert(b > 0);  // 除数必须大于0        BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大        BigInteger m;               // 余数:初始化为0        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = bsearch(b, m);            m -= b*c.s[i];        }        return c.clean();    }    BigInteger operator % (const BigInteger& b) const { //方法与除法相同        BigInteger c = *this;        BigInteger m;        for (int i = s.size()-1; i >= 0; i--) {            m = m*BASE + s[i];            c.s[i] = bsearch(b, m);            m -= b*c.s[i];        }        return m;    }    // 二分法找出满足bx<=m的最大的x    int bsearch(const BigInteger& b, const BigInteger& m) const{        int L = 0, R = BASE-1, x;        while (1) {            x = (L+R)>>1;            if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}            else R = x;        }    }    BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}    BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}    BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}    BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}    BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}    bool operator < (const BigInteger& b) const {        if (s.size() != b.s.size()) return s.size() < b.s.size();        for (int i = s.size()-1; i >= 0; i--)            if (s[i] != b.s[i]) return s[i] < b.s[i];        return false;    }    bool operator >(const BigInteger& b) const{return b < *this;}    bool operator<=(const BigInteger& b) const{return !(b < *this);}    bool operator>=(const BigInteger& b) const{return !(*this < b);}    bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}    bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}};ostream& operator << (ostream& out, const BigInteger& x) {    out << x.s.back();    for (int i = x.s.size()-2; i >= 0; i--) {        char buf[20];        sprintf(buf, "%08d", x.s[i]);        for (int j = 0; j < strlen(buf); j++) out << buf[j];    }    return out;}istream& operator >> (istream& in, BigInteger& x) {    string s;    if (!(in >> s)) return in;    x = s;    return in;}BigInteger ans[51],c[51][51];void init(){    for(int i = 1; i <= 50; ++i)        c[i][0] = c[i][i] = 1;    for(int i = 2; i <= 50; ++i)        for(int j = 1; j < i; ++j)        c[i][j] = c[i-1][j] + c[i-1][j-1];    ans[1] = ans[2] = 1;    for(int i = 3; i <= 50; ++i){        for(int j = 1; j < i; ++j){            long long tmp = (pow(2,j)-1);            ans[i] = ans[i] + ans[j] * ans[i-j]* c[i-2][j-1] * tmp;        }    }}int main(){    init();    int n;    while(scanf("%d", &n) && n){       cout<<ans[n]<<endl;    }}