結果
問題 | No.1124 Earthquake Safety |
ユーザー |
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提出日時 | 2020-07-03 13:42:43 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 411 ms / 3,000 ms |
コード長 | 2,984 bytes |
コンパイル時間 | 979 ms |
コンパイル使用メモリ | 73,600 KB |
実行使用メモリ | 87,808 KB |
最終ジャッジ日時 | 2024-09-16 17:08:36 |
合計ジャッジ時間 | 16,200 ms |
ジャッジサーバーID (参考情報) |
judge6 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 58 |
ソースコード
#include <iostream>#include <vector>using namespace std;#pragma warning (disable: 4996)const long long mod = 1000000007;const long long inv2 = 500000004;// 入出力long long N, A[1 << 19], B[1 << 19];long long Answer1, Answer2, Answer3A, Answer3B, Answer3C;vector<int> X[1 << 19];vector<int> Y[1 << 19];// DPint par[1 << 19];long long dp1[1 << 19];long long dp2[1 << 19];long long dpr[1 << 19][4];long long Eval[1 << 19];void dfs1(int pos) {dp1[pos] += 1LL;for (int i : X[pos]) {if (dp1[i] != 0LL) continue;dfs1(i);par[i] = pos;Y[pos].push_back(i);dp1[pos] += dp1[i] * inv2 % mod;dp1[pos] %= mod;}}void dfs2(int pos) {for (int i : Y[pos]) {dp2[i] += dp2[pos] * inv2 % mod;dp2[i] += dp1[pos] * inv2 % mod;dp2[i] -= dp1[i] * (inv2 * inv2 % mod) % mod;dp2[i] = (dp2[i] + mod * mod) % mod;dfs2(i);}}void getEval() {for (int i = 1; i <= N; i++) {long long r = 0, s1 = 0, s2 = 0;for (int j : Y[i]) s1 += dp1[j] * inv2 % mod;for (int j : Y[i]) s2 += (dp1[j] * inv2 % mod) * (dp1[j] * inv2 % mod) % mod;s1 %= mod; s2 %= mod;r = s1 * s1 - s2;r = (r + mod * mod) % mod;Eval[i] = r * inv2 % mod;}}int main() {// Step #1. 入力scanf("%lld", &N);for (int i = 1; i <= N - 1; i++) scanf("%lld%lld", &A[i], &B[i]);for (int i = 1; i <= N - 1; i++) {X[A[i]].push_back(B[i]);X[B[i]].push_back(A[i]);}// Step #2. 全方位木DPdfs1(1);dfs2(1);// Step #3. [1], [2] の場合を求めるAnswer1 = N;getEval();for (int i = 1; i <= N; i++) Answer2 += (dp1[i] + mod - 1LL) % mod;for (int i = 1; i <= N; i++) Answer2 += Eval[i];Answer2 %= mod;// Step #4. [3] の場合 - Pattern Afor (int i = 1; i <= N; i++) {long long r1 = (dp1[i] - 1LL);long long r2 = dp2[i];Answer3A += r1 * r2 % mod;Answer3A %= mod;}for (int i = 1; i <= N; i++) {Answer3A += Eval[i];Answer3A %= mod;}// Step #5. [3] の場合 - Pattern Bfor (int i = 1; i <= N; i++) {long long r1 = dp2[i];long long r2 = Eval[i];Answer3B += r1 * r2 % mod;Answer3B %= mod;}// Step #6. [3] の場合 - Pattern Cfor (int i = 1; i <= N; i++) {for (int j = 0; j <= Y[i].size(); j++) { dpr[j][0] = 0; dpr[j][1] = 0; dpr[j][2] = 0; dpr[j][3] = 0; }dpr[0][0] = 1;for (int j = 0; j < Y[i].size(); j++) {for (int k = 0; k <= 3; k++) dpr[j + 1][k] = dpr[j][k];for (int k = 0; k <= 2; k++) {dpr[j + 1][k + 1] += dpr[j][k] * (dp1[Y[i][j]] * inv2 % mod) % mod;dpr[j + 1][k + 1] %= mod;}}Answer3C += dpr[Y[i].size()][3];Answer3C %= mod;}// Step #7. 出力long long Ret1 = Answer1;long long Ret2 = Answer2;long long Ret3 = (Answer3A + Answer3B + Answer3C) % mod;for (int i = 1; i <= N - 1; i++) { Ret1 *= 2LL; Ret1 %= mod; }for (int i = 1; i <= N - 1; i++) { Ret2 *= 2LL; Ret2 %= mod; }for (int i = 1; i <= N - 1; i++) { Ret3 *= 2LL; Ret3 %= mod; }cout << (1LL * Ret1 + 6LL * Ret2 + 6LL * Ret3) % mod << endl;return 0;}