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【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());
}
}
}
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