結果

問題 No.2826 Earthwork
コンテスト
ユーザー AngrySadEight
提出日時 2024-07-16 14:48:30
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,766 ms / 5,000 ms
コード長 8,064 bytes
コンパイル時間 25,902 ms
コンパイル使用メモリ 286,552 KB
最終ジャッジ日時 2025-02-23 15:50:32
ジャッジサーバーID
（参考情報） 		judge4 / judge3
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 AC * 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 # 

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <algorithm>
#include <atcoder/all>
#include <bitset>
#include <cassert>
#include <cmath>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using namespace atcoder;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, n) for (int i = (int)(n) - 1; i >= 0; i--)
#define repk(i, k, n) for (int i = k; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define mod1 1000000007
#define mod2 998244353
#define mod3 100000007
#define vi vector<int>
#define vs vector<string>
#define vc vector<char>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvc vector<vector<char>>
#define vvl vector<vector<ll>>
#define vvb vector<vector<bool>>
#define vvvi vector<vector<vector<int>>>
#define vvvl vector<vector<vector<ll>>>
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define vpii vector<pair<int, int>>
#define vpll vector<pair<ll, ll>>
#define vvpii vector<vector<pair<int, int>>>
#define vvpll vector<vector<pair<ll, ll>>>

using mint = modint998244353;

template <typename T>
void debug(T e) {
    cerr << e << endl;
}

template <typename T>
void debug(vector<T> &v) {
    rep(i, v.size()) { cerr << v[i] << " "; }
    cerr << endl;
}

template <typename T>
void debug(vector<vector<T>> &v) {
    rep(i, v.size()) {
        rep(j, v[i].size()) { cerr << v[i][j] << " "; }
        cerr << endl;
    }
}

template <typename T>
void debug(vector<pair<T, T>> &v) {
    rep(i, v.size()) { cerr << v[i].first << " " << v[i].second << endl; }
}

template <typename T>
void debug(set<T> &st) {
    for (auto itr = st.begin(); itr != st.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(multiset<T> &ms) {
    for (auto itr = ms.begin(); itr != ms.end(); itr++) {
        cerr << *itr << " ";
    }
    cerr << endl;
}

template <typename T>
void debug(map<T, T> &mp) {
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
        cerr << itr->first << " " << itr->second << endl;
    }
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << H << " ";
    debug_out(T...);
}

void dijkstra(vvpll &graph, ll v, vl &dist) {
    priority_queue<pll, vpll, greater<pll>> pque;
    pque.push(pll(0, v));
    dist[v] = 0;
    while (pque.size()) {
        pll p = pque.top();
        pque.pop();
        if (dist[p.second] != p.first) {
            continue;
        }
        for (auto x : graph[p.second]) {
            if (dist[x.first] > dist[p.second] + x.second) {
                dist[x.first] = dist[p.second] + x.second;
                pque.push(pll(dist[x.first], x.first));
            }
        }
    }
}

int main() {
    ll N;
    cin >> N;
    vector<vector<ll>> H(N, vector<ll>(N));
    for (ll i = 0; i < N; i++) {
        for (ll j = 0; j < N; j++) {
            cin >> H[i][j];
        }
    }
    vector<string> S(N);
    for (ll i = 0; i < N; i++) {
        cin >> S[i];
    }
    vector<vector<ll>> A(N - 1, vector<ll>(N));
    for (ll i = 0; i < N - 1; i++) {
        for (ll j = 0; j < N; j++) {
            cin >> A[i][j];
        }
    }
    vector<vector<ll>> B(N, vector<ll>(N - 1));
    for (ll i = 0; i < N; i++) {
        for (ll j = 0; j < N - 1; j++) {
            cin >> B[i][j];
        }
    }
    ll Q;
    cin >> Q;
    vector<ll> R(Q);
    vector<ll> C(Q);
    vector<ll> E(Q);
    for (ll i = 0; i < Q; i++) {
        cin >> R[i] >> C[i] >> E[i];
        R[i]--;
        C[i]--;
    }
    vector<vector<pll>> graph1(N * N + 1, vector<pll>(0));  // 順辺のグラフ
    vector<vector<pll>> graph2(N * N + 1, vector<pll>(0));  // 逆辺のグラフ
    for (ll i = 0; i < N - 1; i++) {
        for (ll j = 0; j < N; j++) {
            ll s = H[i][j] + H[i + 1][j];
            if ((i + j) % 2 == 1) {
                graph1[i * N + j].push_back(
                    pll((i + 1) * N + j, A[i][j] * abs(s) - s));
                graph2[(i + 1) * N + j].push_back(
                    pll(i * N + j, A[i][j] * abs(s) - s));
                graph1[(i + 1) * N + j].push_back(
                    pll(i * N + j, A[i][j] * abs(s) + s));
                graph2[i * N + j].push_back(
                    pll((i + 1) * N + j, A[i][j] * abs(s) + s));
            } else {
                graph2[i * N + j].push_back(
                    pll((i + 1) * N + j, A[i][j] * abs(s) - s));
                graph1[(i + 1) * N + j].push_back(
                    pll(i * N + j, A[i][j] * abs(s) - s));
                graph2[(i + 1) * N + j].push_back(
                    pll(i * N + j, A[i][j] * abs(s) + s));
                graph1[i * N + j].push_back(
                    pll((i + 1) * N + j, A[i][j] * abs(s) + s));
            }
        }
    }
    for (ll i = 0; i < N; i++) {
        for (ll j = 0; j < N - 1; j++) {
            ll s = H[i][j] + H[i][j + 1];
            if ((i + j) % 2 == 1) {
                graph1[i * N + j].push_back(
                    pll(i * N + j + 1, B[i][j] * abs(s) - s));
                graph2[i * N + j + 1].push_back(
                    pll(i * N + j, B[i][j] * abs(s) - s));
                graph1[i * N + j + 1].push_back(
                    pll(i * N + j, B[i][j] * abs(s) + s));
                graph2[i * N + j].push_back(
                    pll(i * N + j + 1, B[i][j] * abs(s) + s));
            } else {
                graph2[i * N + j].push_back(
                    pll(i * N + j + 1, B[i][j] * abs(s) - s));
                graph1[i * N + j + 1].push_back(
                    pll(i * N + j, B[i][j] * abs(s) - s));
                graph2[i * N + j + 1].push_back(
                    pll(i * N + j, B[i][j] * abs(s) + s));
                graph1[i * N + j].push_back(
                    pll(i * N + j + 1, B[i][j] * abs(s) + s));
            }
        }
    }
    for (ll i = 0; i < N; i++) {
        for (ll j = 0; j < N; j++) {
            if (S[i][j] == '-') {
                if ((i + j) % 2 == 0) {
                    graph1[N * N].push_back(pll(i * N + j, 0));
                    graph2[i * N + j].push_back(pll(N * N, 0));
                } else {
                    graph2[N * N].push_back(pll(i * N + j, 0));
                    graph1[i * N + j].push_back(pll(N * N, 0));
                }
            } else if (S[i][j] == '+') {
                if ((i + j) % 2 == 1) {
                    graph1[N * N].push_back(pll(i * N + j, 0));
                    graph2[i * N + j].push_back(pll(N * N, 0));
                } else {
                    graph2[N * N].push_back(pll(i * N + j, 0));
                    graph1[i * N + j].push_back(pll(N * N, 0));
                }
            } else if (S[i][j] == '=') {
                graph1[i * N + j].push_back(pll(N * N, 0));
                graph1[N * N].push_back(pll(i * N + j, 0));
                graph2[i * N + j].push_back(pll(N * N, 0));
                graph2[N * N].push_back(pll(i * N + j, 0));
            }
        }
    }
    ll INF = 4000000000000000002;
    vector<ll> dist1(N * N + 1, INF);
    vector<ll> dist2(N * N + 1, INF);
    dist1[N * N] = 0;
    dist2[N * N] = 0;
    dijkstra(graph1, N * N, dist1);
    dijkstra(graph2, N * N, dist2);

    for (ll i = 0; i < Q; i++) {
        if ((R[i] + C[i]) % 2 == 0) {
            if (H[R[i]][C[i]] + dist1[R[i] * N + C[i]] >= E[i]) {
                cout << "Yes" << endl;
            } else {
                cout << "No" << endl;
            }
        } else {
            if (H[R[i]][C[i]] + dist2[R[i] * N + C[i]] >= E[i]) {
                cout << "Yes" << endl;
            } else {
                cout << "No" << endl;
            }
        }
    }
}
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