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2.68 kB
| using namespace std; | |
| const int LIM = 52; | |
| inline int& setmax(int &l, int r) { | |
| return l = max(l, r); | |
| } | |
| int N, K, C[LIM]; | |
| vector<int> adj[LIM]; | |
| // dp[i][j][k] = max value in i's subtree, | |
| // with j new paths present, | |
| // and with a free path ongoing from i's parent if k=1. | |
| int dp[LIM][LIM][2]; | |
| // dp2[i][j][k][c] = max value after first i children of current node, | |
| // with j new paths present, | |
| // and with a free path available for use if k=1, | |
| // and with at least one child connected if c=1. | |
| int dp2[LIM][LIM][2][2]; | |
| void rec(int i, int parent) { | |
| int nc = 0; | |
| for (int j : adj[i]) { | |
| if (j != parent) { | |
| rec(j, i); | |
| nc++; | |
| } | |
| } | |
| // Sub-DP. | |
| for (int j = 0; j <= nc; j++) { | |
| memset(dp2[j], -1, sizeof dp2[j]); | |
| } | |
| dp2[0][0][0][0] = 0; | |
| nc = 0; | |
| for (int j : adj[i]) { | |
| if (j == parent) { | |
| continue; | |
| } | |
| for (int a1 = 0; a1 <= K; a1++) { | |
| for (int x : {0, 1}) { | |
| for (int y : {0, 1}) { | |
| int d = dp2[nc][a1][x][y]; | |
| if (d < 0) { | |
| continue; | |
| } | |
| for (int a2 = 0; a2 <= K - a1; a2++) { | |
| // Connect to child. | |
| int d2 = dp[j][a2][1]; | |
| if (d2 >= 0) { | |
| setmax(dp2[nc + 1][a1 + a2 + (x ? 0 : 1)][1 - x][1], d + d2); | |
| } | |
| // Don't connect to child. | |
| d2 = dp[j][a2][0]; | |
| if (d2 >= 0) { | |
| setmax(dp2[nc + 1][a1 + a2][x][y], d + d2); | |
| } | |
| } | |
| } | |
| } | |
| } | |
| nc++; | |
| } | |
| // Combine into main DP. | |
| memset(dp[i], -1, sizeof dp[i]); | |
| for (int o : {0, 1}) { | |
| for (int j = 0; j <= K; j++) { | |
| for (int x : {0, 1}) { | |
| for (int y : {0, 1}) { | |
| if (!i && !y) { | |
| continue; // Root must connect to at least one child. | |
| } | |
| int d = dp2[nc][j][x][y]; | |
| if (d < 0) { | |
| continue; | |
| } | |
| // Include current node's value? | |
| if (o || y) { | |
| d += C[i]; | |
| } | |
| // Free path ongoing from parent? | |
| setmax(dp[i][j - (o && x ? 1 : 0)][o], d); | |
| } | |
| } | |
| } | |
| } | |
| } | |
| int solve() { | |
| cin >> N >> K; | |
| for (int i = 0; i < N; i++) { | |
| cin >> C[i]; | |
| adj[i].clear(); | |
| } | |
| for (int i = 0; i < N - 1; i++) { | |
| int a, b; | |
| cin >> a >> b; | |
| a--, b--; | |
| adj[a].push_back(b); | |
| adj[b].push_back(a); | |
| } | |
| // DP | |
| rec(0, -1); | |
| int ans = C[0]; | |
| for (int i = 0; i <= K; i++) { | |
| setmax(ans, dp[0][i][0]); | |
| } | |
| return ans; | |
| } | |
| int main() { | |
| int T; | |
| cin >> T; | |
| for (int t = 1; t <= T; t++) { | |
| cout << "Case #" << t << ": " << solve() << endl; | |
| } | |
| return 0; | |
| } | |