結果
問題 | No.1488 Max Score of the Tree |
ユーザー |
|
提出日時 | 2022-03-21 01:48:10 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 32 ms / 2,000 ms |
コード長 | 3,345 bytes |
コンパイル時間 | 4,040 ms |
コンパイル使用メモリ | 232,548 KB |
実行使用メモリ | 42,940 KB |
最終ジャッジ日時 | 2024-10-07 20:22:41 |
合計ジャッジ時間 | 5,336 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 29 |
ソースコード
#include <bits/stdc++.h>#include <atcoder/all>typedef long long int ll;typedef long long int ull;#define MP make_pairusing namespace std;using namespace atcoder;typedef pair<ll, ll> P;const ll MOD = 998244353;// const ll MOD = 1000000007;using mint = modint998244353;// using mint = modint1000000007;const double pi = 3.1415926536;const int MAX = 2000003;long long fac[MAX], finv[MAX], inv[MAX];void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}ll gcd(ll x, ll y) {if (y == 0) return x;else if (y > x) {return gcd (y, x);}else return gcd(x % y, y);}ll lcm (ll x, ll y) {return x / gcd(x, y) * y;}ll my_sqrt(ll x) {ll m = 0;ll M = 3000000001;while (M - m > 1) {ll now = (M + m) / 2;if (now * now <= x) {m = now;}else {M = now;}}return m;}ll keta(ll n) {ll ret = 0;while (n) {n /= 10;ret++;}return ret;}ll ceil(ll n, ll m) {// n > 0, m > 0ll ret = n / m;if (n % m) ret++;return ret;}ll pow_ll(ll x, ll n) {if (n == 0) return 1;if (n % 2) {return pow_ll(x, n - 1) * x;}else {ll tmp = pow_ll(x, n / 2);return tmp * tmp;}}typedef struct {int to;int id;} edge;vector<edge> v[101];int des[101]; // 辺iの子孫の数int cst[101]; // 辺iのコストvoid dfs(int par, int cur, int now_id) {int d = 0;for (auto e : v[cur]) {if (e.to == par) continue;dfs(cur, e.to, e.id);d += des[e.id];}if ((v[cur].size() == 0) || ((v[cur].size() == 1) && (v[cur][0].to == par))) d++;des[now_id] = d;}int main() {int n, k;cin >> n >> k;for (int i = 1; i <= n - 1; i++) {int a, b, c;cin >> a >> b >> c;edge e1, e2;e1.to = b;e1.id = i;cst[i] = c;v[a].push_back(e1);e2.to = a;e2.id = i;v[b].push_back(e2);}dfs(0, 1, 0);// for (int i = 1; i <= n - 1; i++) {// cout << des[i] << ' ' << cst[i] << endl;// }int dp[n + 1][k + 1]; // i番目まででj長さ時の最大追加料for (int i = 0; i <= n; i++) {for (int j = 0; j <= k; j++) {dp[i][j] = 0;}}int first_ans = 0;for (int i = 1; i <= n - 1; i++) {first_ans += cst[i] * des[i];}// cout << first_ans << endl;for (int i = 0; i < n; i++) {for (int j = 0; j <= k; j++) {dp[i + 1][j] = max(dp[i][j], dp[i + 1][j]); // 追加しないif (j + cst[i + 1] <= k) {dp[i + 1][j + cst[i + 1]] = max(dp[i + 1][j + cst[i + 1]], dp[i][j] + des[i + 1] * cst[i + 1]);}}}int add = 0;for (int i = 0; i <= k; i++) {add = max(add, dp[n][i]);}cout << first_ans + add << endl;return 0;}