POJ 2486 Apple Tree (DP)

#include <iostream> #include <cstring> #include <deque>using namespace std; #define SIZE 230#define Back 1#define away 0int dp[size][size][2];bool visits[size];int vals[size];d eque< int > Tree[size];int num, steps;v    OID dfs (int u) {Visits[u] = true;    const int len = Tree[u].size ();    for (int i = 0; I <= steps; ++i) dp[u][i][back] = Dp[u][i][away] = Vals[u];        for (int i = 0; i < len; ++i) {int son = tree[u][i];        if (Visits[son]) continue;        DFS (son);                    for (int s = steps; s >= 0,--s) {for (int ss = 0; SS <= S; ++ss) {/*                Starting from U, go back to u, need to walk two more steps u->son,son->u, assign to son Tree SS step, other subtree s-ss step, all return. */Dp[u][s + 2][back] = max (Dp[u][s + 2][back], Dp[u][s-ss][bac                K] + dp[son][ss][back]);                /* Do not back u (go to other subtrees of U), return at son. */                Dp[u][s + 2][away] = max (Dp[u][s + 2][away], Dp[u][s-ss][away]                + Dp[son][ss][back]);                /* Go through the other subtrees of U, go back to u, traverse son subtree, and go one more step in the current subtree. */Dp[u][s + 1][away] = max (Dp[u][s + 1][away], Dp[u][s-ss][bac            K] + dp[son][ss][away]);    }}}}int Main () {int u, v;        while (CIN >> num >> steps) {memset (DP, 0, sizeof (DP));        Memset (visits, false, sizeof (visits));        for (int i = 1; I <= num; ++i) tree[i].clear ();        for (int i = 1; I <= num; ++i) cin >> Vals[i];            for (int i = 1; I <= num-1; ++i) {cin >> u >> v;            Tree[u].push_back (v);        Tree[v].push_back (U);        } dfs (1); cout << Max (Dp[1][steps][back], Dp[1][steps][away]) << Endl; } return 0;}

POJ 2486 Apple Tree (DP)