【Codeforces 467C】George and Job

【链接】 我是链接,点我呀:)

【题意】

让你从1..n这n个数字中
选出来k个不相交的长度为m的区间
然后这个k个区间的和最大
求出这k个区间的和的最大值

【题解】

设dp[i][j]表示前i个数字已经选出了j个区间的最大值
看看是以当前位置为结尾选择一个区间,还是这个位置不包括在任何一个区间里.
分这两种情况转移就好

【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 = 5000;
    static class Task{

        int n,m,k;
        int a[];
        long sum[];
        long dp[][];

        long get_sum(int l,int r) {
        	return sum[r]-sum[l-1];
        }

        public void solve(InputReader in,PrintWriter out) {
        	n = in.nextInt();m = in.nextInt();k = in.nextInt();
        	a = new int[N+10];
        	sum = new long[N+10];
        	dp = new long[N+10][N+10];

        	for (int i = 1;i <= n;i++) {
        		a[i] = in.nextInt();
        	}
        	for (int i = 1;i <= n;i++) sum[i]=sum[i-1]+a[i];
        	long ans = 0;
        	dp[0][0] = 0;
        	for (int i = 1;i <= n;i++)
        		for (int j = 0;j <= Math.min(k, i);j++) {
        			if (j==0) {
        				dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);
        			}else {
        				if (i-m>=0) {
        					dp[i][j] = Math.max(dp[i][j], dp[i-m][j-1]+get_sum(i-m+1,i));
        					dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);
        				}
        			}
        			ans = Math.max(ans, dp[i][j]);
        		}
        	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());
        }
    }
}