CF 23E (Tree-backpack merge)

Problem 2 tree (tree. cpp/c/pas) [Title Description] L invented a tree-related game (Friendly reminder: the tree is a connected graph without loops ): he deletes any number of edges (which can be 0) from the tree, calculates the product of all connected block sizes after deletion, and L gets so many scores. Your task is to find the maximum score L can get for a given tree. [Input format] the first line is an integer n, indicating the number of nodes in the tree. The next n-1 rows have two integers a [I], B [I] (1 <= a [I], B [I] <= n ), the edge connecting a [I] and B [I. Make sure that the input graph is a tree. Output Format: returns an integer that represents the maximum score that can be obtained by L. [Example input] Example 1: 51 22 33 44 5 Example 2: 81 21 32 42 53 63 76 8 Example 3: 31 21 3 [sample output] Example 1: 6 Example 2: 18 Example 3: 3 [data range] for 10% of data, 1 <= n <= 5; for 30% of data, 1 <= n <= 100; and 30% of data, ensure that data is a chain. For 100% of data, 1 <= n <= 700; f [I] [j] indicates that the connected block of the father of I has the maximum value of j in subtree I. So this is the combination of the tree-like Dp + backpack, and the combination of the backpack, and then the third ...... [Cpp] # include <cstdio> # include <cstring> # include <algorithm> # include <functional> # include <iostream> # include <cstdlib> # include <cmath> using namespace std; # define MAXN (700 + 10) # define ll long # define F (100000000) int n, edge [MAXN * 2], pre [MAXN], next [MAXN * 2], size = 0, son [MAXN]; struct bign {ll a [40]; bign () {memset (a, 0, sizeof (a); a [0] = 1;} bign (int x) {memset (a, 0, sizeof (a); a [0] = 1; a [1] = x;} ll & op Erator [] (const int I) {return a [I];} friend bign operator * (bign a, bign B) {bign c; for (int I = 1; I <= a [0]; I ++) for (int j = 1; j <= B [0]; j ++) {c [I + J-1] + = a [I] * B [j]; if (c [I + J-1]> F) {c [I + j] + = c [I + J-1]/F; c [I + J-1] % = F ;}} c [0] = min (a [0] + B [0], (long) 39); while (! C [c [0] & c [0]> 1) c [0] --; return c;} friend bool operator> (bign a, bign B) {if (a [0]! = B [0]) return a [0]> B [0]; for (int I = a [0], j = B [0]; I> 0; I --, j --) if (a [I]! = B [j]) return a [I]> B [j]; return false;} void print () {printf ("% I64d ", a [a [0]); for (int I = a [0]-1; I --) {printf ("%. 8I64d ", a [I]) ;}} f [MAXN] [MAXN]; bign max (bign a, bign B) {if (a> B) return; return B;} void addedge (int u, int v) {edge [++ size] = v; next [size] = pre [u]; pre [u] = size;} void dfs (int x, int father) {son [x] = 1; // f [x] [1] = 1; f [x] [1] = 1; for (int p = pre [x]; p = next [p]) {int & v = edge [p]; if (V! = Father) {dfs (v, x);/* for (int I = son [x] + son [v]; I> 0; I --) {if (I <son [x]) f [x] [I] = f [x] [I] * son [v]; for (int k = son [v]; k> = 0; k-) if (i-k-1> = 0) f [x] [I] = max (f [x] [I], f [x] [i-k-1] * f [v] [k]);} son [x] + = son [v]; bign maxv = son [v]; for (int k = 0; k <= son [v]-1; k ++) maxv = max (maxv, f [v] [k] * (k + 1); f [x] [0] = f [x] [0] * maxv; */for (int I = son [x]; I --) for (int j = son [v]; j> = 0; j --) f [x] [I + j] = max (f [x] [I + j], f [x] [I] * f [v] [j]); son [x] + = son [v] ;}} f [x] [0] = bign (son [x]); for (int I = 1; I <= son [x]; I ++) f [x] [0] = max (f [x] [0], f [x] [I] * bign (I); return;} int main () {// freopen ("tree. in "," r ", stdin); // freopen (" tree. out "," w ", stdout); scanf (" % d ", & n); memset (pre, 0, sizeof (pre); memset (next, 0, sizeof (next); for (int I = 1; I <n; I ++) {int u, v; scanf ("% d", & u, & v); addedge (u, v); addedge (v, u);} addedge (n + 1, 1); dfs (n + 1, 0 ); // n + 1 is ans/* for (int I = 1; I <= n + 1; I ++) {for (int j = 0; j <= son [I]-1; j ++) {f [I] [j]. print (); printf ("");} printf ("\ n");} * // f [n + 1] [1] = bign (123456789) * bigns (234567899); f [1] [0]. print (); printf ("\ n"); return 0 ;}