Codeforces Round #267 (Div. 2) C. George and Job(DP)

C. George and Job

题意：在序列 $a_i$中选出 $k$个不相交大小为 $m$的区间，使其区间和最大。
题解： $dp[i][j]$表示在前 $j$个数里选 $i$个区间的最大区间和。则有 $dp[i][j] = max(dp[i - 1][j - m] + sum[j] - sum[j - m], dp[i][j - 1])$

代码

#include<bits/stdc++.h>

#define DEBUG(x) std::cerr << #x << '=' << x << std::endl
typedef long long LL;
using namespace std;
const int N = 5024;
int a[N];
LL sum[N], dp[N][N];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0);
	int n,m,k;
	cin >> n >> m >> k;
	for(int i = 1; i <= n; ++i) {
		cin >> a[i];
		sum[i] = sum[i - 1] + a[i];
	}
	for(int i = 1; i <= k; ++i) {
		for(int j = m; j <= n; ++j) 
			dp[i][j] = max(dp[i - 1][j - m] + sum[j] - sum[j - m], dp[i][j - 1]);
	}
	cout << dp[k][n] << endl;
    return 0;
}