hdu1294 Rooted Trees Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1294

　　组合数学的题，求给定结点数可以构造出多少不同的根树。

　　我的做法跟网上其他的做法有点不同，我是用整数划分的方法将一种结点数的子结点全部分解，然后用乘法原理求出组合数是多少。刚开始的时候忘记处理重复子树的会出现交换以后相同的状况，然后通过推理，发现了重复的个数跟组合数有关，所以开始的时候可以用杨辉三角构造组合数。

　　例如，如果有n个结点的子树种类有k种，同时有这样的m棵子树，那么这几棵子树的组合数就是Σ((k - i) * C(i + m - 2, m - 2))(0 <= i < k)。然后通过一个深度优先搜索每一种划分的情况，最后就可以得出组合的总数了。

　　代码如下；

#include <cstdio>
#include <cstring>
#include <cstdlib>

typedef __int64 ll;

ll lw, hh;
const int mod = 1000000000;
ll rc[41], sm;
ll c[300][21];
bool pr;

void pre(){
    memset(c, 0, sizeof(c));
    c[0][0] = 1;
    for (int i = 1; i < 300; i++){
        c[i][0] = 1;
        for (int j = 1; j < 21; j++){
            c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
        }
    }
}

ll cal(ll a, int n){
    ll ret = 0;

    if (a == 1 || n == 1) return a;
    if (n == 2) return (a + 1) * a / 2;
    if (n == 3){
        for (ll i = 1; i <= a; i++){
            ret += i * (a + 1 - i);
        }

        return ret;
    }
    for (ll i = 0; i < a; i++){
        ret += c[i + n - 2][n - 2] * (a - i);
    }

    return ret;
}

int dv[41];

void dfs(int rest, int last, int pos){
    if (!rest){
        ll rt = 1, ls;
        int cnt;

        ls = dv[0];
        dv[pos] = 0;
        cnt = 1;
        for (int i = 1; i <= pos; i++){
            if (dv[i] != ls){
                rt *= cal(rc[ls], cnt);
                cnt = 1;
                ls = dv[i];
            }
            else cnt++;
            //if (pr) printf("%d ", dv[i - 1]);
        }
        //if (pr) printf("rt  %I64d\n", rt);
        sm += rt;
        return ;
    }
    if (last > rest) last = rest;
    for (int i = last; i > 0; i--){
        dv[pos] = i;
        dfs(rest - i, i, pos + 1);
    }
}

void print(){
    for (int i = 0; i <= 40; i++){
    }
    rc[1] = rc[2] = 1;
    rc[3] = 2;
    for (int i = 4; i <= 40; i++){
        if (i == 13) pr = true;
        else pr = false;
        sm = 0;
        dfs(i - 1, i - 1, 0);
        rc[i] = sm;
#ifndef ONLINE_JUDGE
        printf(" %I64d,", sm);
        puts("");
#endif
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in", "w", stdout);
#endif
    pre();
    print();

#ifdef ONLINE_JUDGE
    int n;
    while (~scanf("%d", &n))
        printf("%I64d\n", rc[n]);
#endif

    return 0;
}

yuan神的做法：http://www.cppblog.com/Yuan/archive/2010/10/22/130929.html

　　模仿yuan神打了一个代码，关键的地方在于重复集合的组合：

　　引用：由于子树不分顺序，所以枚举出来的相同规模的子树是一种有重集的组合 ，用公式C(n+m-1,m)

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>

using namespace std;

typedef __int64 ll;
ll ans[41], cnt[41];

ll combi(ll n, ll m){
    ll ret = 1;

    if (m > n - m) m = n - m;
    for (ll i = 0; i < m; i++){
        ret *= n - i;
        ret /= i + 1;
    }

    return ret;
}

void dfs(int n, int cur, int rest){
    if (!rest){
        ll pro = 1;

        for (int i = 1; i <= n; i++){
            if (cnt[i])
                pro *= combi(ans[i] - 1 + cnt[i], cnt[i]);
        }
        ans[n] += pro;

        return ;
    }
    if (cur > rest) return ;
    cnt[cur]++;
    dfs(n, cur, rest - cur);
    cnt[cur]--;
    dfs(n, cur + 1, rest);
}

void pre(){
    ans[1] = ans[2] = 1;
    for (int i = 3; i <= 40; i++){
        dfs(i, 1, i - 1);
    }
}

int main(){
    pre();
    int n;

    while (cin >> n){
        cout << ans[n] << endl;
    }

    return 0;
}

——written by Lyon