UVA 10253Series-Parallel Networks

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题意：给定n，求有n条边的串并联网络的方案数。

分析：《算法竞赛入门经典训练指南》数学基础例题7。书中给出的建模思想很好，并且对组合数学求解和dp的边界情况分析都很好。多学习，多积累。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=1000010;const int MAX=151;const int MOD1=1000007;const int MOD2=100000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000009;const ll INF=10000000010;typedef double db;typedef unsigned long long ull;ll f[35],dp[35][35];ll C(ll n,ll m) {    db ans=1;    for (int i=1;i<=m;i++) ans*=n-i+1;    for (int i=1;i<=m;i++) ans/=i;    return (ll)(ans+0.5);}int main(){    int i,j,k,n=30;    f[1]=1;    memset(dp,0,sizeof(dp));    for (i=0;i<=n;i++) dp[i][0]=1;    for (i=1;i<=n;i++) { dp[0][i]=0;dp[i][1]=1; }    for (i=1;i<=n;i++) {        for (j=2;j<=n;j++) {            dp[i][j]=0;            for (k=0;k*i<=j;k++)            dp[i][j]+=C(f[i]+k-1,k)*dp[i-1][j-k*i];        }        f[i+1]=dp[i][i+1];    }    while (scanf("%d", &n)&&n)    printf("%lld\n", n==1 ? 1:2*f[n]);    return 0;}

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