CodeForces 1992D Test of Love

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-16

4.CodeForces 1992D Test of Love2024-07-16

5.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)2024-07-2817.CodeForces 908B New Year and Buggy Bot2024-08-0118.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 1992D【Test of Love】

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思路

       从起点开始起跳，找出下一个木头的位置，若与当前位置的距离小于等于m，则可以直接跳过去，否则判断当前位置与下一个木头之间有没有鳄鱼，有鳄鱼则不能到达对岸，否则继续查找下一个木头，直到对岸。

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代码

#include <functional>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stdio.h>
#include <string>
#include <cstring>
#include <vector>
using namespace std;
#define ll long long
const int N = 500 + 10;
 
void solve(){
  int n, m, k;
  cin >> n >> m >> k;
  string s;
  cin >> s;
  s = "L" + s + "L";
  n += 2;
  int cur = 0, ans = 0;
  while (cur < n - 1) {
    int p = s.find("L", cur + 1);
    if (p == -1)
      break;
    if (p - cur <= m) {
      cur = p;
      continue;
    }
    cur += m;
    string tmp = s.substr(cur, p - cur);
    if (tmp != string(p - cur, 'W'))
      break;
    ans += p - cur;
      cur=p;
   }
   if (cur < n - 1 || ans > k)
     cout << "NO\n";
   else
     cout << "YES\n";
}
 
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}