【Codeforces 225C】Barcode

【链接】 我是链接,点我呀:)
【题意】

让你把每一列都染成一样的颜色
要求连续相同颜色的列的长度都大于等于x小于等于y
问你最少的染色次数

【题解】

先求出每一列染成#或者.需要染色多少次
设f[0][i][j]表示前i列,以i为结尾的连续列长度为j的#列最少需要染色多少次
设f[1][i][j]表示前i列,以i为结尾的连续列长度为j的.列最少需要染色多少次
f[0][i][j]可以由f[0][i-1][j-1]转移过来
但是j==1的时候比较特殊
会由f[1][i-1][x..y]转移过来(由里面的最小值转移)
为了方便
所以在算f[1][i-1][x..y]的时候
可以把这些的最小值存在f[0][i][0]里面
这样就不用每次转移j的时候都重新求最小值了(不过临时求问题也不大)
最后取min(f[0][m][x..y],f[1][m][x..y])

【代码】

import java.io.*;
import java.util.*;

public class Main {
    
    
    static InputReader in;
    static PrintWriter out;
        
    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }
    
    static int N = (int)1e3;
    static class Task{
        String []s;
        int cost[][];
        int f[][][];
        
        public void solve(InputReader in,PrintWriter out) {
            s = new String[N+10];
            cost = new int[2][N+10];
            f = new int[2][N+10][N+10];
            int n,m,x,y;
            n = in.nextInt();m = in.nextInt();x = in.nextInt();y = in.nextInt();
            for (int i = 0;i <= n-1;i++)  s[i] = in.next();
            for (int i = 0;i < n;i++)
                for (int j = 0;j < m;j++) {
                    char key = s[i].charAt(j);
                    if (key=='#')
                        cost[0][j+1]++;
                }
            for (int j = 1;j <= m;j++) cost[1][j] = n-cost[0][j];
            for (int p = 0;p < 2;p++)
                for (int i = 0;i <= N;i++)
                    for (int j = 0;j <= N;j++)
                        f[p][i][j] = (int)1e7;
            f[0][1][1] = cost[0][1];
            if (x<=1 && 1<=y) f[1][1][0] = cost[0][1];
            
            f[1][1][1] = cost[1][1];
            if (x<=1 && 1<=y) f[0][1][0] = cost[1][1];
            
            for (int i = 2;i <= m;i++)
                for (int j = 1;j <= y;j++){
                        f[0][i][j] = Math.min(f[0][i][j], f[0][i-1][j-1]+cost[0][i]);
                        if (x<=j && j<=y) f[1][i][0] = Math.min(f[1][i][0], f[0][i][j]);
                        
                        f[1][i][j] = Math.min(f[1][i][j], f[1][i-1][j-1]+cost[1][i]);
                        if (x<=j && j<=y) f[0][i][0] = Math.min(f[0][i][0], f[1][i][j]);
                        
                        //f[0][i][1] = min{f[1][i-1][1],f[1][i-1][2],f[1][i-1][3]...f[1][i-1][y]
                        //f[1][i][1] = min(f[0][i-1][1],f[0][i-1][2],f[0][i-1][3]...f[0][i-1][y]
                    }
            int ans = (int)1e7;
            for (int i = x;i <= y;i++) {
                ans = Math.min(ans, f[1][m][i]);
                ans = Math.min(ans, f[0][m][i]);
            }
            out.println(ans);
        }
    }

    

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;
        
        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }
        
        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}