hdu 1294 rooted trees problem

Rooted Trees Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 337    Accepted Submission(s): 108

Problem Description

Give you two definitions tree and rooted tree. An undirected connected graph without cycles is called a tree. A tree is called rooted if it has a distinguished vertex r called the root. Your task is to make a program to calculate the number of rooted trees with n vertices denoted as Tn. The case n=5 is shown in Fig. 1.

 

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer means n(1<=n<=40) .

 

Output

For each test case, there is only one integer means Tn.

 

Sample Input

125

 

Sample Output

119

 

Author

SmallBeer(CML)

 

Source

杭电ACM集训队训练赛（VIII）

 

Recommend

lcy

分情况，分为有 1, 2, 3, ..., n-1 个子树的情况，然后相加。

#include <cstdio>#include <cstdlib>#include <cassert>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;long long GCD (long long a, long long b) {    return b ? GCD (b, a % b) : a;}long long C (long long n, long long m) {    long long num = 1; // numerator    long long don = 1; // doneminator    if (n - m < m) {        m = n - m;    }    long long gcd;    while (m) {        gcd = GCD (num, don);        num /= gcd;        don /= gcd;        num *= n--;        don *= m--;    }    assert (num % don == 0);    return num / don;}int a [10000][50]; // once this array is accessed violation, variable sz's value may be modified!!!int sz;void Devide (int last, int len, int rest, int ind) {#ifdef _DEBUG    //printf ("Devide (last = %d, len = %d, rest = %d, ind = %d)\n", last, len, rest, ind);#endif    if (ind == len) {        assert (sz + 1 < 10000); // this assertion can catch the bug.        assert (rest == 0);        for (int i=0; i<len; ++i) {            a [sz + 1][i] = a [sz][i];        }        ++sz;        return;    }    assert (sz < 100000); // this assertion cannot catch the bug, because ...#ifdef _DEBUG    int t = lrint (ceil (1.0 * rest / (len - ind)));    assert (t * (len - ind) >= rest);    assert ((t - 1) * (len - ind) < rest);#endif    for (int i=lrint(ceil(1.0*rest/(len-ind))); i<=last && len-ind-1<=rest-i; ++i) {        a [sz][ind] = i;        Devide (i, len, rest-i, ind+1);    }}void Make (int n, int len) {#ifdef _DEBUG    //printf ("Make (n = %d, len = %d)\n", n, len);#endif    sz = 0;#ifdef _DEBUG    int t = lrint (ceil (1.0 * n / len));    assert (t * len >= n);    assert ((t - 1) * len < n);#endif    for (int i=lrint(ceil(1.0*n/len)); i<=n-len+1; ++i) {        a [sz][0] = i;        Devide (i, len, n-i, 1);    }#ifdef _DEBUG    for (int i=0; i<sz; ++i) {        for (int k=0; k<len; ++k) {            printf ("%d%s", a [i][k], k==len-1 ? "\n" : " ");        }    }#endif}long long dp [50];long long DP (long long n) { // it's no longer a dp!!! what it is named is just for historic reason.#ifdef _DEBUG    //printf ("DP (n = %lld)\n", n);#endif    for (int i=1; i<n; ++i) {        sz = 0;        Make (n - 1, i);        for (int s=0; s<sz; ++s) {            long long t = 1;            int k = 0, m = 0;            while (k < i && m < i) {                m = k;                while (++m < i && a [s][m] == a [s][k]) {}                t *= C (dp [a [s][k]] + m - k - 1, m - k);                k = m;            }            dp [n] += t;#ifdef _DEBUG            printf ("dp [%lld] = %lld\n", n, dp [n]);#endif        }    }    return dp [n];}void MakeDP () {    memset (dp, 0, sizeof (dp));    dp [1] = dp [2] = 1;    for (int i=3; i<=40; ++i) {        DP (i);    }}int main () {    MakeDP ();    int n;    while (scanf ("%d", &n) == 1) {        printf ("%lld\n", dp [n]);    }    return 0;}