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The two-fork tree of n nodes is the nth item of Catalan number
For a subtrees tree, the moved node is a definite Location. So just to know the number of K nodes that have been determined where there is an empty child pointer m, and the node in the N indeterminate position under that subtree equals to the number of species of the m-nullable subtree constructed with n nodes. The number of shapes for the whole tree is the product of several individual subtree morphology numbers.
Define dp[i][j] to construct the number of forms of J tree with I Nodes. dp[i][j] = sum{dp[i-1][j-k] * catalan[k] | 0≤k≤j}, The DP array is preprocessed with the complexity of O (n^3). Then the simulation operation calculates the M and N of each Subtree.
#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef long ll; const int MAXN = 505;const int mod = 1e9 + 7;int dp[maxn][maxn], catalan[maxn];void Preserve () {catalan[0] = catalan[1] = 1;for (int i = 2; i <=; i++) {for (int j = 0; J < i; j + +) catalan[i] = (catalan[i] + (ll) catalan[j] * catalan[i -j-1])% mod;} dp[0][0] = 1;for (int i = 1; i <=; i++) {for (int j = 0; J <=; J + +) {for (int k = 0; k <= j; k++) dp[i][j] = (dp[i][j] + (ll) dp[i-1][j-k] * catalan[k])% mod;}}} int N, M, far[maxn], son[maxn][2], idx[maxn], cnt[maxn], sum[maxn];inline int newNode (int F) {m++;cnt[m] = son[m][0] = So n[m][1] = 0;far[m] = F;return M;} void init () {M = 0;int u = newNode (1), t, k;idx[u] = m;for (int i = 1; i <= N; i++) {scanf ("%d", &t); if (t = = 0) u = Far[u];else if (t <= 2) {if (son[u][t-1] = = 0) {son[u][t-1] = newNode (u); idx[son[u][t-1]] = idx[u];cnt[idx[u]]--;} U = son[u][t-1];} else {scanf ("%d", &k); son[u][t-3] = NewNode (u); cnt[son[u][t-3]] = k-1;idx[son[u][t-3]] = son[u][t-3];}} memset (sum, 0, sizeof (sum)), for (int i = 1; i <= M; i++) {if (son[i][0] = 0) sum[idx[i]]++;if (son[i][1] = = 0) Sum[idx[i ]]++;}} int solve () {int ret = 1;for (int i = 1; i <= M; i++) {if (idx[i]! = i) Continue;ret = (ll) ret * dp[sum[i]][cnt[i]]% mod;} Return ret;} int main () {preserve (); int cas = 1;while (scanf ("%d", &n) = = 1) {init ();p rintf ("case #%d:%d\n", cas++, solve ());} Return 0;}
finally, a large number of processing methods to calculate the results after the submission WA crunch Leng thought for a long while do not know why crunch finally found the last HIMDD blog post, found to be able to use DP Ideas. Solution one: DP Code and parsing such as the following: #include <iost
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