結果

問題 No.1124 Earthquake Safety
ユーザー e869120
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
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ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <vector>
using namespace std;
#pragma warning (disable: 4996)
const long long mod = 1000000007;
const long long inv2 = 500000004;
//
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];
// DP
int 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.
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. DP
dfs1(1);
dfs2(1);
// Step #3. [1], [2]
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] - Pattern A
for (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] - Pattern B
for (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] - Pattern C
for (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.
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;
}
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