#687294 (C++14) No.1638 Robot Maze

• 問題: No.1638 Robot Maze
• ユーザー: tnakao0123
• 提出日時: 2021-08-07 18:50:33
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 4 ms / 2,000 ms
• コード長: 1,870 bytes
• コンパイル時間: 941 ms
• コンパイル使用メモリ: 99,824 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-09-18 19:36:59
• 合計ジャッジ時間: 2,417 ms
• ジャッジサーバーID （参考情報）: judge2 / judge4

ファイルパターン	結果
sample	AC * 3
other	AC * 49

ソースコード

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/* -*- coding: utf-8 -*-
 *
 * 1638.cc:  No.1638 Robot Maze - yukicoder
 */
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;
/* constant */
const int MAX_H = 100;
const int MAX_W = 100;
const int MAX_N = MAX_H * MAX_W;
const long long LINF = 1LL << 62;
const int dys[] = { -1, 1, 0, 0 }, dxs[] = { 0, 0, 1, -1 };
enum { DU, DD, DR, DL };
/* typedef */
typedef long long ll;
typedef pair<ll,int> pli;
/* global variables */
char fs[MAX_H][MAX_W + 4];
ll ds[MAX_N];
/* subroutines */
inline int yx2p(int y, int x, int w) { return y * w + x; }
inline void p2yx(int p, int &y, int &x, int w) { y = p / w, x = p % w; }
/* main */
int main() {
  int h, w, dcs[4], p, sy, sx, gy, gx;
  ll k;
  scanf("%d%d%d%d%d%d%lld%d%d%d%d%d",
    &h, &w, dcs + DU, dcs + DD, dcs + DR, dcs + DL, &k, &p,
    &sy, &sx, &gy, &gx);
  sy--, sx--, gy--, gx--;
  int st = yx2p(sy, sx, w), gl = yx2p(gy, gx, w);
  for (int y = 0; y < h; y++) scanf("%s", fs[y]);
  fill(ds, ds + h * w, LINF);
  ds[st] = 0;
  priority_queue<pli> q;
  q.push(pli(0, st));
  while (! q.empty()) {
    pli u = q.top(); q.pop();
    ll ud = -u.first;
    int up = u.second;
    if (up == gl) break;
    int uy, ux;
    p2yx(up, uy, ux, w);
    for (int di = 0; di < 4; di++) {
      int vy = uy + dys[di], vx = ux + dxs[di];
      if (vy >= 0 && vy < h && vx >= 0 && vx < w && fs[vy][vx] != '#') {
    int vp = yx2p(vy, vx, w);
    ll vd = ud + dcs[di] + (fs[vy][vx] == '@' ? p : 0);
    if (ds[vp] > vd) {
      ds[vp] = vd;
      q.push(pli(-vd, vp));
    }
      }
    }
  }
  if (ds[gl] <= k) puts("Yes");
  else puts("No");
  return 0;
}
 
 
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