제출 #332950

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
332950		KWang31	수열 (APIO14_sequence)	Java	71 / 100	645 ms	131072 KiB

sequence

//Convex Hull Trick, verified on CF319C (319 is ID)
import java.io.*; import java.util.*;
public class sequence{
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        
    } 
    
    static int[] M;
    static long[] B;
    static int[] ind;
    static int pointer=0;
    static short[][] ch; static boolean[][] b1; static boolean[][] b2;
    static int N;
    static int sz;
    public static void main(String[] args){
        
        FastReader br=new FastReader();
        N=br.nextInt(); int K=br.nextInt();
        int[] s=new int[N]; 
        long[] dp=new long[N];
        long[] dp2=new long[N];
        ch=new short[N][K]; b1=new boolean[N][K];b2=new boolean[N][K];
        s[0]=br.nextInt(); for(int i=1;i<N;i++){s[i]=s[i-1]+br.nextInt(); }
        //System.out.println(Arrays.toString(s));
        for (int i = 0; i < K; i++) {
            M=new int[N];
            B=new long[N];
            ind=new int[N]; sz=0;
            add(s[0],-(long)s[0]*s[0],i);
            for (int j = 1; j < N; j++) {
                dp2[j]=query((long)s[j],N*i+j);
                add(s[j], dp[j]-(long)s[j]*s[j],Math.max(j,i));
                //System.out.println(M);
                //System.out.println(B);
            }
            
            for (int j = 0; j < N; j++) {
                dp[j]=dp2[j];
            }
            //System.out.println(Arrays.toString(dp));
        }
        //System.out.println(Arrays.toString(dp));
        System.out.println(dp[N-1]);
        StringBuilder sb=new StringBuilder();
        int[] arl=new int[K];
        int cur=N-1;
        for (int i = K-1; i >=0; i--) {
            
            if(b1[cur][i] && b2[cur][i])cur=(int) 4*ch[cur][i]+3;
            else if(b1[cur][i]) cur=(int) 4*ch[cur][i]+1;
            else if(b2[cur][i]) cur=(int) 4*ch[cur][i]+2;
            else cur=(int) 4*ch[cur][i];
            arl[i]=cur+1; 
        }
        //StringBuilder is actually quite memory consuming
        for (int i = 0; i <K; i++) {
            sb.append(arl[i]+" ");
        }
        System.out.println(sb.toString());
    }
    public static boolean bad(int l1, int l2, int l3){//Using doubles to prevent overflow
        return (M[l1]-M[l3])*((double)B[l2]-B[l1])>=(M[l1]-M[l2])*((double)B[l3]-B[l1]);
        //Intersection of l1 and l3 must be ABOVE l2
        //For l2 to be added
        //l1.m>l2.m>l3.m, as sum is strictly increasing
    }
    public static void add(int m, long b, int i){
        M[sz]=m; B[sz]=b; ind[sz]=i; sz++;
        while(sz>=3 && bad(sz-3,sz-2,sz-1)){
            M[sz-2]=m; B[sz-2]=b; ind[sz-2]=i; sz--;
        }
    }
    public static long query(long x, int j){
        if(pointer>=sz){pointer=sz-1;}
        while(pointer<sz-1 && 
                M[pointer+1]*x+B[pointer+1] > M[pointer]*x+B[pointer]){
            pointer++;
        }
        
        ch[j%N][j/N]=(short)(ind[pointer]/4);
        if(ind[pointer]%2==1){
            b1[j%N][j/N]=true;
        }
        if(ind[pointer]%4>=2){
            b2[j%N][j/N]=true;
        }
        return M[pointer]*x+B[pointer];
    }
    
    
}
//Debugging:
//Are you sure your algorithm is correct?
//Bounds: long
//Special cases: n=0,1?
//Make sure you remove your debugging code before you submit!

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 0 / 29