CodeForces 908B New Year and Buggy Bot

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-164.CodeForces 1992D Test of Love2024-07-165.CodeForces 1992B Angry Monk2024-07-166.CodeForces 1992A Only Pluses2024-07-167.CodeForces 1983B Corner Twist2024-07-168.CodeForces 1983A Array Divisibility2024-07-169.CodeForces 1983C Have Your Cake and Eat It Too2024-07-1610.CodeForces 1883B Chemistry2024-07-2611.CodeForces 1883A Morning2024-07-2612.CodeForces 1883C Raspberries2024-07-2813.CodeForces 1883D In Love2024-07-2814.CodeForces 1883E Look Back2024-07-2815.CodeForces 1883F You Are So Beautiful2024-07-2816.CodeForces 1883G1 Dances (Easy version)2024-07-28

17.CodeForces 908B New Year and Buggy Bot2024-08-01

18.CodeForces 908C New Year and Curling2024-08-0119.CodeForces 1132B Discounts2024-08-0120.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 908B【New Year and Buggy Bot】

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思路

       简单模拟，用pair数组存下四个方向然后，依次枚举全排列，将每个方向依次映射给0，1，2，3，然后就是跟着String走，遇到障碍或者走出地图就返回false，表示当前方案是错误的，走完String的所有指示时还没有找到出口也返回false。

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代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 60;
 
int n, m, len, res;
vector<pair<int, int>> ve;
pair<int, int> start, target;
char mp[N][N];
string s;
 
bool check(int x, int y) {
  if (x >= 1 && y >= 1 && x <= n && y <= m && mp[x][y] != '#')
    return true;
  else
    return false;
}
 
bool search() {
  pair<int, int> now = start;
  for (int i = 1; i <= len; i++) {
    now.first += ve[s[i] - '0'].first, now.second += ve[s[i] - '0'].second;
    if (check(now.first, now.second)) {
      if (now == target) {
        return true;
      }
    } else {
      return false;
    }
  }
  return false;
}
 
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%s", mp[i] + 1);
    for (int j = 1; j <= m; j++) {
      if (mp[i][j] == 'S') {
        start.first = i, start.second = j;
      } else if (mp[i][j] == 'E') {
        target.first = i, target.second = j;
      }
    }
  }
  cin >> s;
  len = s.length();
  s = " " + s;
  
  ve.push_back({0, 1});
  ve.push_back({1, 0});
  ve.push_back({0, -1});
  ve.push_back({-1, 0});
  sort(ve.begin(), ve.end());
  
  do {
    if (search()) {
      res++;
    }
  } while (next_permutation(ve.begin(), ve.end()));
 
  cout << res << endl;
  return;
}
 
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
 
  return 0;
}