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

　　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：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 typedef long long LL;
 9 const int N = 33;
10 LL ans[N];
11 int s[N], top;
12 
13 LL com(LL n, LL m) {
14     LL ret = 1;
15     for (int i = 0; i < m; i++) ret *= n - i, ret /= i + 1;
16     return ret;
17 }
18 
19 void dfs(int r, int x) {
20     if (r <= 0) {
21         if (top <= 1) return ;
22         int cnt = 1;
23         LL tmp = 1;
24         for (int i = 1; i <= top; i++) {
25             if (s[i] == s[i + 1]) cnt++;
26             else {
27                 tmp *= com(ans[s[i]] + cnt - 1, cnt);
28                 cnt = 1;
29             }
30         }
31         ans[x] += tmp;
32         return ;
33     }
34     for (int i = s[top]; i <= r; i++) {
35         s[++top] = i, s[top + 1] = -1;
36         dfs(r - i, x);
37         top--;
38     }
39 }
40 
41 void PRE() {
42     ans[1] = 1;
43     for (int i = 2; i < N; i++) {
44         s[top = 0] = 1;
45         ans[i] = 0;
46         dfs(i, i);
47         //cout << i << ' ' << ans[i] << endl;
48     }
49     for (int i = 2; i < N; i++) ans[i] <<= 1;
50 }
51 
52 int main() {
53     PRE();
54     int n;
55     while (cin >> n && n) cout << ans[n] << endl;
56     return 0;
57 }

View Code

UPD：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 typedef long long LL;
 9 const int N = 32;
10 LL dp[N][N], ans[N];
11 
12 LL com(LL n, LL m) {
13     LL ret = 1;
14     for (int i = 0; i < m; i++) {
15         ret *= n - i;
16         ret /= i + 1;
17     }
18     return ret;
19 }
20 
21 void PRE() {
22     ans[1] = 1;
23     dp[0][0] = 1;
24     for (int i = 1; i < N; i++) {
25         dp[i][0] = dp[i][1] = 1;
26         for (int j = 2; j < N; j++) {
27             dp[i][j] = 0;
28             for (int k = 0; k * i <= j; k++) {
29                 dp[i][j] += com(ans[i] + k - 1, k) * dp[i - 1][j - k * i];
30             }
31         }
32         ans[i + 1] = dp[i][i + 1];
33         //cout << ans[i + 1] << endl;
34     }
35 }
36 
37 int main() {
38     PRE();
39     int n;
40     while (cin >> n && n) cout << (n > 1 ? ans[n] << 1 : 1) << endl;
41     return 0;
42 }

View Code

——written by Lyon

posted @ 2013-07-28 02:18 LyonLys 阅读(277) 评论(0) 收藏举报

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