【Codeforces 711C】Coloring Trees

【链接】 我是链接,点我呀:)
【题意】

连续相同的数字分为一段 你可以改变其中0为1~m中的某个数字(改变成不同数字需要不同花费) 问你最后如果要求分成恰好k段的话,最少需要多少花费

【题解】

dp[i][j][k]前i棵树,分成j段,第j段最后一棵树颜色为m的最小花费 很好转移了,分情况就好

【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 = 100;
    static class Task{
        
        int n,m,k;
        long dp[][][] = new long[N+10][N+10][N+10];
        int p[][] = new int[N+10][N+10];
        int c[] = new int[N+10];
        //dp[i][j][k]前i棵树,分成j段,第j段最后一棵树颜色为m的最小花费
        
        public void solve(InputReader in,PrintWriter out) {
        	n = in.nextInt(); m = in.nextInt();k = in.nextInt();
        	for (int i = 1;i <= n;i++) c[i] = in.nextInt();
        	for (int i = 1;i <= n;i++)
        		for (int j = 1;j <= m;j++)
        			p[i][j] = in.nextInt();
        	for (int i = 0;i <=N;i++)
        		for (int j = 0;j <= N;j++)
        			for(int k = 0;k <=N;k++)
        				dp[i][j][k] = -1;
        	if (c[1]==0) {
        		for (int j = 1;j <= m;j++) {
        			dp[1][1][j] = p[1][j];
        		}
        	}else {
        		dp[1][1][c[1]] = 0;
        	}
        	for (int i = 1;i < n;i++)
        		for (int j = 1;j <= k;j++)
        			for (int l = 1;l <= m;l++) {
        				if (dp[i][j][l]!=-1) {
        					if(c[i+1]==0) {
        						for (int l2=1;l2<=m;l2++) {
        							long temp = dp[i][j][l] + p[i+1][l2];
        							if (l2==l){
        								if (dp[i+1][j][l2]==-1) {
        									dp[i+1][j][l2] = temp;
        								}else {
        									dp[i+1][j][l2] = Math.min(dp[i+1][j][l2],temp);
        								}
        							}else {
        								if (dp[i+1][j+1][l2]==-1) {
        									dp[i+1][j+1][l2] = temp;
        								}else {
        									dp[i+1][j+1][l2] = Math.min(dp[i+1][j+1][l2], temp);
        								}
        							}
        						}
        					}else {
        						//c[i+1]=x
        						if (c[i+1]==l) {
        							if (dp[i+1][j][l]==-1) {
        								dp[i+1][j][l] = dp[i][j][l];
        							}else {
        								dp[i+1][j][l] = Math.min(dp[i+1][j][l], dp[i][j][l]);
        							}
        						}else {
        							if (dp[i+1][j+1][c[i+1]]==-1) {
        								dp[i+1][j+1][c[i+1]] = dp[i][j][l];
        							}else {
        								dp[i+1][j+1][c[i+1]] = Math.min(dp[i+1][j+1][c[i+1]], dp[i][j][l]);
        							}
        						}
        					}
        				}
        			}
        	//dp[n][k][1~m]
        	long ma = -1;
        	for (int i = 1;i <= m;i++)
        		if (dp[n][k][i]!=-1) {
        			if (ma==-1)
        				ma = dp[n][k][i];
        			else
        				ma = Math.min(ma, dp[n][k][i]);
        		}
        	out.println(ma);
        }
    }

    

    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());
        }
    }
}