結果
問題 | No.867 避難経路 |
ユーザー |
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提出日時 | 2023-05-01 23:04:03 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2,034 ms / 6,000 ms |
コード長 | 2,641 bytes |
コンパイル時間 | 2,631 ms |
コンパイル使用メモリ | 216,060 KB |
最終ジャッジ日時 | 2025-02-12 16:29:16 |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 41 |
ソースコード
#include <bits/stdc++.h>//#include <atcoder/all>using namespace std;//using namespace atcoder;//using mint = modint1000000007;//const int mod = 1000000007;//using mint = modint998244353;//const int mod = 998244353;//const int INF = 1e9;const long long LINF = 1e18;#define rep(i, n) for (int i = 0; i < (n); ++i)#define rep2(i,l,r)for(int i=(l);i<(r);++i)#define rrep(i, n) for (int i = (n-1); i >= 0; --i)#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)#define all(x) (x).begin(),(x).end()#define allR(x) (x).rbegin(),(x).rend()#define endl "\n"#define P pair<long long,pair<int,int>>template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }const int dx[] = { 1,0,-1,0 };const int dy[] = { 0,-1,0,1 };int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int h, w; cin >> h >> w;int gx, gy; cin >> gx >> gy;gx--, gy--;vector a(h, vector<int>(w));rep(i, h)rep(j, w)cin >> a[i][j];vector dp(250, vector(h, vector<long long>(w, LINF)));rep(i, 250) {long long k = i;dp[i][gx][gy] = k * k + a[gx][gy];priority_queue<P, vector<P>, greater<P>> q;q.push({ dp[i][gx][gy], {gx, gy} });while (!q.empty()) {auto tmp = q.top();q.pop();long long cost = tmp.first;int x = tmp.second.first;int y = tmp.second.second;if (cost > dp[i][x][y]) continue;rep(j, 4) {int nx = x + dx[j];int ny = y + dy[j];if ((nx < 0) || (nx >= h) || (ny < 0) || (ny >= w))continue;long long ncost = cost + k * k + a[nx][ny];if (chmin(dp[i][nx][ny], ncost))q.push({ ncost, { nx, ny } });}}}vector dp2(h, vector<long long>(w, LINF));{set<pair<int, int>>st;dp2[gx][gy] = a[gx][gy];st.insert({ gx,gy });while (!st.empty()) {set<pair<int, int>>st2;for (auto e : st) {int x = e.first;int y = e.second;rep(j, 4) {int nx = x + dx[j];int ny = y + dy[j];if ((nx < 0) || (nx >= h) || (ny < 0) || (ny >= w))continue;if (LINF == dp2[nx][ny])st2.insert({ nx, ny });if (!st2.count({ nx,ny })) continue;long long ncost = dp2[x][y] + a[nx][ny];chmin(dp2[nx][ny],ncost);}}swap(st2, st);}}int q; cin >> q;while (q--) {int x, y; cin >> x >> y;x--, y--;long long k; cin >> k;if (k >= 250) {long long ans = dp2[x][y];long long d = abs(gx - x) + abs(gy - y) + 1;ans += k * k * d;cout << ans << endl;}else {cout << dp[k][x][y] << endl;}}return 0;}