CF467C. George and Job（前綴和+DP）

George and Job

題意：給定一個數組，從中選擇k個長爲m的子段（不重疊），使和最大
分析：
f(i,j) 前i個數中選j個長爲m的子段
轉移方程：當前i不選，繼續沿用i-1的值 或者 選最後m個組成一個子序列
$dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+sum[i]-sum[i-m])$

Code:

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a));
#define lowbit(x) (x & -x)
#define lrt nl, mid, rt << 1
#define rrt mid + 1, nr, rt << 1 | 1
template <typename T>
inline void read(T& t) {
    t = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        t = t * 10 + ch - '0';
        ch = getchar();
    }
    t *= f;
}
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const ll Inf = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x7f7f7f7f;
const double eps = 1e-5;
const double Pi = acos(-1);
const int maxn = 5e3 + 5;

int an[maxn];
ll sum[maxn], dp[maxn][maxn];

int main(void) {
    int n, m, k;
    read(n), read(m), read(k);
    for (int i = 1; i <= n; i++) {
        read(an[i]);
        sum[i] = sum[i - 1] + (ll)an[i];
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= k; j++) {
            dp[i][j] = dp[i - 1][j];
            if (i - m >= 0)
                dp[i][j] =
                    max(dp[i][j], dp[i - m][j - 1] + sum[i] - sum[i - m]);
        }
    }
    printf("%lld\n", dp[n][k]);
    return 0;
}