標籤：方法 current color 公眾號 length hdu tor 推薦 pretty
pid=5370">題目連結：hdu 5370 Tree Maker
n個節點的二叉樹種類為Catalan數的第n項
對於一棵子樹而言，被移動過的節點就是確定的位置。所以僅僅要知道已經確定位置的K個節點有多少個空孩子指標M，和就該子樹下的N個未確定位置的節點，等於是說用N個節點構造M個可為空白的子樹的種類數。對於整個樹的形態數即為若干棵獨立的子樹形態數的乘積。
定義dp[i][j]為用i個節點構造j棵樹的形態數。dp[i][j] = sum{ dp[i-1][j-k] * catalan[k] | 0 ≤ k ≤j }， 用o(n^3)的複雜度預先處理出dp數組。然後類比操作後計算出每一個子樹的M和N。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long 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 <= 500; 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 <= 500; i++) {for (int j = 0; j <= 500; 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] = son[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 ();printf("Case #%d: %d\n", cas++, solve ());}return 0;}
最後用大數的處理方式計算出了結果 提交後WA嗄 愣是想了半天不知道為何嗄最後的最後 找了himdd的博文，發現能夠用DP的思想。解法一：DP代碼及解析例如以下：#include<iost
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