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;
}