Algorithm: zoj-3626 treasure Hunt I (tree dp)

Meaning

to an N-node tree, the node number 1~n, each node has the right value val[i], this value can be obtained through this node (cannot be repeated)

Each side has a cost value of W (i,j), which means that the edges of the I and J points will cost W (i,j).

Now to start with the K point, the total cost is M, ask the total cost of not more than m in the case and ultimately to return to the starting point, how much value can be obtained?

Ideas

Simple tree DP.

F (i,j) is the value of the subtree I, which can be obtained with cost J

Pairs with each son of I, which can be assigned to spend 2*w, 2*w+1, 2*w+2,... J to it, can be seen as a group of items

Pack a backpack for all the sons.

F (i, j) = max{max{f (i, j-k) + f (V, k-2*w) | 2*w<=k<=i} | v is son of I node}

Ans = f (k, m);

Code

/**===================================================== * is a solution for ACM/ICPC problem * * @source: Zoj -3626 Treasure Hunt I * @description: Tree DP * @author: Shuangde * @blog: blog.csdn.net/shuangde800 * @email: Zengshuang
de@gmail.com * Copyright (C) 2013/08/26 16:23 All rights reserved. *======================================================*/#include <iostream> #include <cstdio> # Include <algorithm> #include <vector> #include <queue> #include <cmath> #include <cstring
    
> #include <string> #include <map> #include <set> #define MP make_pair using namespace std;
typedef pair<int, int >PII;
typedef long long Int64;
Const double PI = ACOs (-1.0);
    
const int INF = 0X3F3F3F3F;
const int MAXN = 110;
int N, K, m;
vector<pii>adj[maxn];
int VAL[MAXN];
BOOL VIS[MAXN];
    
int F[MAXN][2*MAXN];
    
void Dfs (int u) {Vis[u] = true;
    
for (int i = 0; I <= m ++i) f[u][i] = Val[u]; for (iNT e = 0; E < adj[u].size ();
++e) {int v = adj[u][e].first int w = adj[u][e].second; if (vis[v]) continue; Dfs (v); for (int i = m; I >= 0;-i) {
for (int j = 2*w J <= i; ++j) {F[u][i] = max (F[u][i], f[u][i-j] + f[v][j-2*w);} int main () {while (~SCANF ("%d", &n)) {//init//www.bianceng.cn for (int i = 0; I <= N; ++i) A
    
Dj[i].clear ();
    
for (int i = 1; I <= n; ++i) scanf ("%d", &val[i)); for (int i = 0; i < n-1 ++i) {int u, V, W; scanf ("%d%d%d", &u, &v, &w); Adj[u].push_back (MP (V, W));
Dj[v].push_back (MP (U, W));
} scanf ("%d%d", &k, &m);
memset (f, 0, sizeof (f));
memset (Vis, 0, sizeof (VIS));
DFS (k);
printf ("%d\n", F[k][m]);
return 0; }