結果

問題 No.2826 Earthwork
ユーザー achapi
提出日時 2024-07-26 10:26:57
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
RE  
実行時間 -
コード長 2,222 bytes
コンパイル時間 2,845 ms
コンパイル使用メモリ 214,784 KB
最終ジャッジ日時 2025-02-23 18:07:27
ジャッジサーバーID
(参考情報)
judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other RE * 40
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#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(), 1e18);
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<pair<int, long long>>> G(N * N + 1);
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) % 2 * 2);
G[i * N + j].push_back(make_pair((i + 1) * N + j, abs(T) * A[i][j] + T));
G[(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) % 2 * 2);
G[i * N + j].push_back(make_pair(i * N + j + 1, abs(T) * B[i][j] + T));
G[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) % 2) == 0){
G[i * N + j].push_back(make_pair(N * N, 0));
}
if (S[i][j] == '=' or ((S[i][j] == '-') ^ (i + j) % 2) == 1){
G[N * N].push_back(make_pair(i * N + j, 0));
}
}
}
auto D = Dijkstra(G, N * N);
int Q;
cin >> Q;
while (Q--){
int R, C, E;
cin >> R >> C >> E;
R--;
C--;
if (H[R][C] + D[R * N + C] >= E){
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}
}
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