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 = ;
 LL ans[N];
 int s[N], top;

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

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

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

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

UPD：

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

 using namespace std;

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

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

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

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

——written by Lyon