結果
問題 | No.2826 Earthwork |
ユーザー | achapi |
提出日時 | 2024-07-26 11:26:50 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 1,500 ms / 5,000 ms |
コード長 | 2,425 bytes |
コンパイル時間 | 2,886 ms |
コンパイル使用メモリ | 229,028 KB |
実行使用メモリ | 157,648 KB |
最終ジャッジ日時 | 2024-07-26 11:27:25 |
合計ジャッジ時間 | 34,771 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,816 KB |
testcase_01 | AC | 42 ms
6,940 KB |
testcase_02 | AC | 1,361 ms
156,980 KB |
testcase_03 | AC | 148 ms
17,624 KB |
testcase_04 | AC | 1,104 ms
133,324 KB |
testcase_05 | AC | 257 ms
36,096 KB |
testcase_06 | AC | 682 ms
85,336 KB |
testcase_07 | AC | 893 ms
107,324 KB |
testcase_08 | AC | 1,322 ms
157,024 KB |
testcase_09 | AC | 1,366 ms
157,360 KB |
testcase_10 | AC | 1,412 ms
157,096 KB |
testcase_11 | AC | 1,012 ms
156,708 KB |
testcase_12 | AC | 41 ms
6,940 KB |
testcase_13 | AC | 448 ms
51,636 KB |
testcase_14 | AC | 1,500 ms
157,604 KB |
testcase_15 | AC | 320 ms
41,820 KB |
testcase_16 | AC | 619 ms
86,004 KB |
testcase_17 | AC | 200 ms
25,936 KB |
testcase_18 | AC | 536 ms
83,096 KB |
testcase_19 | AC | 239 ms
36,092 KB |
testcase_20 | AC | 1,099 ms
129,380 KB |
testcase_21 | AC | 42 ms
6,944 KB |
testcase_22 | AC | 888 ms
104,040 KB |
testcase_23 | AC | 860 ms
102,600 KB |
testcase_24 | AC | 1,349 ms
157,148 KB |
testcase_25 | AC | 532 ms
107,280 KB |
testcase_26 | AC | 845 ms
104,824 KB |
testcase_27 | AC | 28 ms
6,944 KB |
testcase_28 | AC | 1,423 ms
157,648 KB |
testcase_29 | AC | 102 ms
15,592 KB |
testcase_30 | AC | 58 ms
8,480 KB |
testcase_31 | AC | 97 ms
14,092 KB |
testcase_32 | AC | 35 ms
6,940 KB |
testcase_33 | AC | 752 ms
86,496 KB |
testcase_34 | AC | 1,367 ms
157,020 KB |
testcase_35 | AC | 312 ms
39,988 KB |
testcase_36 | AC | 1,412 ms
157,600 KB |
testcase_37 | AC | 1,431 ms
157,376 KB |
testcase_38 | AC | 668 ms
79,892 KB |
testcase_39 | AC | 65 ms
9,908 KB |
testcase_40 | AC | 52 ms
7,576 KB |
ソースコード
#include <bits/stdc++.h> using namespace std;template<class T> vector<T> Dijkstra(vector<vector<pair<int, T>>> &G, int s){ vector<T> D(G.size(), 4e18); D[s] = 0; priority_queue<pair<T, int>, vector<pair<T, int>>, greater<pair<T, int>>> q; q.push(make_pair(0, s)); while (!q.empty()){ auto [d, u] = q.top(); q.pop(); if (D[u] < d){ continue; } for (auto [v, c] : G[u]){ d = D[u] + c; if (d < D[v]){ D[v] = d; q.push(make_pair(d, v)); } } } return D; } int main(){ int N; cin >> N; vector<vector<int>> H(N, vector<int>(N)); for (int i = 0; i < N; i++){ for (int j = 0; j < N; j++){ cin >> H[i][j]; } } vector<string> S(N); for (int i = 0; i < N; i++){ cin >> S[i]; } vector<vector<long long>> A(N - 1, vector<long long>(N)); for (int i = 0; i < N - 1; i++){ for (int j = 0; j < N; j++){ cin >> A[i][j]; } } vector<vector<long long>> B(N, vector<long long>(N - 1)); for (int i = 0; i < N; i++){ for (int j = 0; j < N - 1; j++){ cin >> B[i][j]; } } vector<vector<vector<pair<int, long long>>>> G(2, vector<vector<pair<int, long long>>>(N * N + 1)); vector<vector<long long>> D(2); for (int k = 0; k < 2; k++){ for (int i = 0; i < N - 1; i++){ for (int j = 0; j < N; j++){ long long T = (H[i][j] + H[i + 1][j]) * (1 - (i + j + k) % 2 * 2); G[k][i * N + j].push_back(make_pair((i + 1) * N + j, abs(T) * A[i][j] + T)); G[k][(i + 1) * N + j].push_back(make_pair(i * N + j, abs(T) * A[i][j] - T)); } } for (int i = 0; i < N; i++){ for (int j = 0; j < N - 1; j++){ long long T = (H[i][j] + H[i][j + 1]) * (1 - (i + j + k) % 2 * 2); G[k][i * N + j].push_back(make_pair(i * N + j + 1, abs(T) * B[i][j] + T)); G[k][i * N + j + 1].push_back(make_pair(i * N + j, abs(T) * B[i][j] - T)); } } for (int i = 0; i < N; i++){ for (int j = 0; j < N; j++){ if (S[i][j] == '?'){ continue; } if (S[i][j] == '=' or ((S[i][j] == '-') ^ (i + j + k) % 2) == 0){ G[k][i * N + j].push_back(make_pair(N * N, 0)); } if (S[i][j] == '=' or ((S[i][j] == '-') ^ (i + j + k) % 2) == 1){ G[k][N * N].push_back(make_pair(i * N + j, 0)); } } } D[k] = Dijkstra(G[k], N * N); } int Q; cin >> Q; while (Q--){ int R, C; long long E; cin >> R >> C >> E; R--; C--; if (D[(R + C) % 2][R * N + C] + H[R][C] >= E){ cout << "Yes" << endl; } else { cout << "No" << endl; } } }