POJ - 1737 Connected Graph（计数dp）

一时失言乱红尘 2021-11-29 05:26 236阅读 0赞

## 题意：求n个点组成的连通图的种数。 ##

## 分析：经典DP。详见《算法竞赛进阶指南》P337。 ##

## 代码： ##

#include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #define ll long long
    using namespace std;
    const int N = 60, S = 600;
    int n;
    struct A {
    	int a[S], len;
    	inline A operator / (const int x) const {
    		A ans;
    		memset(ans.a, 0, sizeof(ans.a));
    		ans.len = 0;
    		int num = 0;
    		for (int i = len; i; i--) {
    			num = num * 10 + a[i];
    			ans.a[i] = num / x;
    			num %= x;
    			if (!ans.len && ans.a[i]) ans.len = i;
    		}
    		return ans;
    	}
    	inline A operator + (const A x) const {
    		A ans;
    		memset(ans.a, 0, sizeof(ans.a));
    		for (int i = 1; i <= max(len, x.len); i++) {
    			ans.a[i] += a[i] + x.a[i];
    			ans.a[i+1] = ans.a[i] / 10;
    			ans.a[i] %= 10;
    		}
    		ans.len = max(len, x.len);
    		if (ans.a[ans.len+1]) ++ans.len;
    		return ans;
    	}
    	inline A operator * (const A x) const {
    		A ans;
    		memset(ans.a, 0, sizeof(ans.a));
    		for (int i = 1; i <= len; i++)
    			for (int j = 1; j <= x.len; j++) {
    				ans.a[i+j-1] += a[i] * x.a[j];
    				ans.a[i+j] += ans.a[i+j-1] / 10;
    				ans.a[i+j-1] %= 10;
    			}
    		ans.len = len + x.len - 1;
    		if (ans.a[ans.len+1]) ++ans.len;
    		return ans;
    	}
    } f[N], p[N];
    
    inline A C(int x, int y) {
    	A ans;
    	ans.len = ans.a[1] = 1;
    	for (int i = y, j = 1; j <= x; i--, j++) {
    		int t = i;
    		A tmp;
    		tmp.len = 0;
    		while (t) {
    			tmp.a[++tmp.len] = t % 10;
    			t /= 10;
    		}
    		ans = ans * tmp / j;
    	}
    	return ans;
    }
    
    inline void print(A x) {
    	for (int i = x.len; i; i--) printf("%d", x.a[i]);
    	puts("");
    }
    
    int main() {
    	for (int i = 1; i <= 50; i++) {
    		ll t = (1ll << i) - 1;
    		while (t) {
    			p[i].a[++p[i].len] = t % 10;
    			t /= 10;
    		}
    	}
    	f[1].len = f[2].len = f[1].a[1] = f[2].a[1] = 1;
    	for (int i = 3; i <= 50; i++)
    		for (int j = 1; j <= i - 1; j++)
    			f[i] = f[i] + C(j - 1, i - 2) * f[j] * f[i-j] * p[j];
    	while (cin >> n && n) print(f[n]);
    	return 0;
    }