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CF830A/831D Office Keys

思路:

问题的关键在于对钥匙按照位置排序之后,最终选择的n个钥匙一定是其中的一个连续的区间。

实现:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 const ll INF = 0x3f3f3f3f3f3f3f3f;
 8 ll man[1005], key[2005];
 9 ll cal(ll m, ll k, ll p)
10 {
11     if ((k >= m && k <= p) || (k <= m && k >= p)) return abs(m - p);
12     return abs(m - k) + abs(k - p);
13 }
14 int main()
15 {
16     ll n, k, p, ans = INF;
17     cin >> n >> k >> p;
18     for (int i = 0; i < n; i++) cin >> man[i];
19     for (int i = 0; i < k; i++) cin >> key[i];
20     sort(man, man + n); sort(key, key + k);
21     for (int i = 0; i <= k - n; i++)
22     {
23         ll maxn = 0;
24         for (int j = 0; j < n; j++)
25         {
26             maxn = max(maxn, cal(man[j], key[j + i], p));
27         }
28         ans = min(ans, maxn);
29     }
30     cout << ans << endl;
31     return 0;
32 }

 

CF830A/831D Office Keys