CodeForces-1256E Yet Another Division

传送门：CodeForces-1256E

思路分析

要使极差最小，肯定相近的在一个队，先对数组排序
每个队伍的人数至少3个最多5个，因为6个以上可以分为两个队伍
设dp[i]表示前i个人的最小代价
枚举每一个左端点，以及队伍长度，如果能更新答案，则将左端点的位置记录在右端点上

样例输入

6
1 5 12 13 2 15

样例输出

7 2
2 2 1 1 2 1

AC代码

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
using namespace __gnu_pbds;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;

const int INF = 0x3f3f3f3f;
const int N= 2e5+10;

int dp[N],l[N],ans[N];
pair<int,int>a[N];

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    read(n);
    for(int i=1; i<=n; i++) read(a[i].fi),a[i].se=i;
    mem(dp,INF);
    dp[0]=0;
    sort(a+1,a+1+n);
    for(int i=1; i<=n; i++) {
        for(int j=3; j<=5 && j+i-1<=n; j++) {
            int cost=dp[i-1]+a[i+j-1].fi-a[i].fi;
            if(cost<dp[i+j-1]) {
                dp[i+j-1]=cost;
                l[i+j-1]=i;
            }
        }
    }

    int now=n;
    int cnt=0;
    while(now) {
        ++cnt;
        for(int i=l[now]; i<=now; i++) ans[a[i].se]=cnt;
        now=l[now]-1;

    }
    printf("%d %d\n",dp[n],cnt);
    for(int i=1; i<=n; i++) printf("%d ",ans[i]);




    return 0;
}