uva 10253 Series-Parallel Networks （整数划分+多重集）

UVa Online Judge

　　题意是计算给定数量的边通过串联并联两种方式，能组成多少种不同的网络。将它转化为一个树形结构，也就是求有多少不同构的树。

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;
const int N = 33;
LL ans[N];
int s[N], top;

LL com(LL n, LL m) {
    LL ret = 1;
    for (int i = 0; i < m; i++) ret *= n - i, ret /= i + 1;
    return ret;
}

void dfs(int r, int x) {
    if (r <= 0) {
        if (top <= 1) return ;
        int cnt = 1;
        LL tmp = 1;
        for (int i = 1; i <= top; i++) {
            if (s[i] == s[i + 1]) cnt++;
            else {
                tmp *= com(ans[s[i]] + cnt - 1, cnt);
                cnt = 1;
            }
        }
        ans[x] += tmp;
        return ;
    }
    for (int i = s[top]; i <= r; i++) {
        s[++top] = i, s[top + 1] = -1;
        dfs(r - i, x);
        top--;
    }
}

void PRE() {
    ans[1] = 1;
    for (int i = 2; i < N; i++) {
        s[top = 0] = 1;
        ans[i] = 0;
        dfs(i, i);
        //cout << i << ' ' << ans[i] << endl;
    }
    for (int i = 2; i < N; i++) ans[i] <<= 1;
}

int main() {
    PRE();
    int n;
    while (cin >> n && n) cout << ans[n] << endl;
    return 0;
}

UPD：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;
const int N = 32;
LL dp[N][N], ans[N];

LL com(LL n, LL m) {
    LL ret = 1;
    for (int i = 0; i < m; i++) {
        ret *= n - i;
        ret /= i + 1;
    }
    return ret;
}

void PRE() {
    ans[1] = 1;
    dp[0][0] = 1;
    for (int i = 1; i < N; i++) {
        dp[i][0] = dp[i][1] = 1;
        for (int j = 2; j < N; j++) {
            dp[i][j] = 0;
            for (int k = 0; k * i <= j; k++) {
                dp[i][j] += com(ans[i] + k - 1, k) * dp[i - 1][j - k * i];
            }
        }
        ans[i + 1] = dp[i][i + 1];
        //cout << ans[i + 1] << endl;
    }
}

int main() {
    PRE();
    int n;
    while (cin >> n && n) cout << (n > 1 ? ans[n] << 1 : 1) << endl;
    return 0;
}

——written by Lyon