题目描述
在一根数轴上,你需要处理n个目标物,你只有k个炸弹,每个炸弹可以你随意放置在x的地方,它的爆炸范围是这个区间,你需要求解k个炸弹的情况下最小的r是多少?
Solution
滑动窗口,二分求解。目标物的下标不一定有序,我们直接对目标物的下标进行排序。
紧接着二分这个答案r,使用一个类似与滑动窗口的办法去处理炸弹数目,最后判断是否大于k个炸弹数即可。
#pragma GCC target("avx,sse2,sse3,sse4,popcnt") #pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math") #include <bits/stdc++.h> using namespace std; #define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0) #define all(__vv__) (__vv__).begin(), (__vv__).end() #define endl "\n" #define pai pair<int, int> #define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__)) typedef long long ll; typedef unsigned long long ull; typedef long double ld; inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; } inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op) putchar(op); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); if (op) putchar(op); } inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; } inline int lowbit(int x) { return x & (-x); } const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} }; const int MOD = 1e9 + 7; const int INF = 0x3f3f3f3f; const int N = 5e4 + 7; int a[N], n, k; bool check(int x) { int cnt = 1, last = a[1] + 2 * x; for (int i = 2; i <= n; ++i) { if (a[i] > last) { ++cnt; last = a[i] + 2 * x; } if (cnt > k) return false; } return true; } int main() { n = read(), k = read(); for (int i = 1; i <= n; ++i) a[i] = read(); sort(a + 1, a + 1 + n); int l = 1, r = 1e9 + 10, ans; while (l <= r) { int mid = l + r >> 1; if (check(mid)) ans = mid, r = mid - 1; else l = mid + 1; } print(ans); return 0; }