hihocoder #1400 : Composition 2017 Microsoft Autumn Recruitment (dp)

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#1400 : Composition

analyze

It should not be difficult to think of DP

in characters$c$The maximum length that can be saved at the end must be the optimal solution that can be saved at the end of the previous character

Assume

$cnt[c]$: in characters$c$The maximum string length that can be stored at the end.

then iterate over the entire string, for each character in it$e$

$cnt[e] = max\{cnt[e],cnt[c]+1\} c\in ['a'-'z']$

java code

import java.io.BufferedInputStream;
import java.nio.MappedByteBuffer;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Stack;



public class Main {
    public static final int INF = 0x3f3f3f3f;
    public static Scanner in = new Scanner(new BufferedInputStream(System.in));
    public static void main(String[] args) {
        String string;
        boolean[][] map = new boolean[27][27];
        for(int i='a' ; i<='z' ; ++i)
            for(int j='a' ; j<='z' ; ++j)map[i-'a'][j-'a'] = true;
        in.nextLine();
        string = in.nextLine();
        while (string=="")string = in.nextLine();
        int k = in.nextInt();
        while (k-->0) {
            String string2 = in.nextLine();
            while (string2.equals(""))string2 = in.nextLine();
            int u = string2.charAt(0)-'a';
            int v = string2.charAt(1) - 'a';
            map[u][v] = map[v][u] = false;
        }
        System.out.println(new Solution(string, map).solve());
    }
}
class Solution {
    String string;
    boolean[][] map;
    int[] cnt;
    public Solution(String string, boolean[][] map) {
        super();
        this.string = string;
        this.map = map;
        cnt = new int['z'+10];
        for(int i='a' ; i<='z' ; ++i)cnt[i] =0;
    }

    public int solve() {
        for(int i=0 ; i<string.length() ; ++i){
            char now = string.charAt(i);
            int tmp =0;
            for(int j='a' ; j<='z' ; ++j){
                if(map[now-'a'][j-'a'])tmp = Math.max(tmp, cnt[j]+1);
            }
            cnt[now] = Math.max(tmp, cnt[now]);
        }
        int ret =0;
        for(int i='a' ; i<'z' ; ++i)ret = Math.max(ret, cnt[i]);
        return string.length()-ret;
    }
}