CF-883

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I. Photo Processing

题意

将 n 个数分堆，要保证每堆不少于 k 个数，每堆的值为这堆数中的最大值减去最小值，取这些值中的最大值，要是这个最大值最小。

题解

二分答案，$dp$ 判断是否可行。$dp[i]$ 表示将 $1\sim i$ 划分成满足条件的区间，能划分最远的地方。

代码

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#include <bits/stdc++.h>
using namespace std;
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define fro1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define fro0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define oper(type) bool operator <(const type y)const
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<long long, long long> pll;
typedef vector<long long> vll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
typedef vector<int> vii;
typedef long double db;
typedef long long ll;
typedef int itn;
int in(int &a,int &b,int &c,int &d){return scanf("%d%d%d%d",&a,&b,&c,&d);}
int in(int &a,int &b,int &c){return scanf("%d%d%d",&a,&b,&c);}
int in(int &a,int &b){return scanf("%d%d",&a,&b);}
int in(ll &a){return scanf("%lld",&a);}
int in(int &a){return scanf("%d",&a);}
int in(char *s){return scanf("%s",s);}
int in(char &s){return scanf("%c",&s);}
int in(db &a){return scanf("%Lf",&a);}
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const db eps = 1e-8;
const int N = 3.1e5;
const ll mod = 1e9+7;
int sign(db a) { return a < -eps ? -1 : a > eps; }
int db_cmp(db a, db b){ return sign(a-b); }

int a[N],n,k,dp[N];
bool check(int x){
    int last=0;
    forl(i,k,n){
        int j=dp[i-k];
        if(a[i]-a[j+1]<=x)last=i;
        dp[i]=last;
    }
    return dp[n]==n;
}
int main() {
#ifdef PerpEternal
    // freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    in(n,k);
    for1(i,n)in(a[i]);
    sort(a+1,a+1+n);
    int l=0,r=1e9,mid;
    while(l<=r){
        mid=(l+r)/2;
        if(check(mid))r=mid-1;
        else l=mid+1;
    }
    printf("%d\n",l);
    return 0;
}