HDU - 1024 DP

Max Sum Plus Plus

Topic meaning:

• Give you n numbers, let you find m consecutive and disjoint intervals from them, and then make the sum of all interval elements to be the largest.

data range:

$1\leq n\leq 10^{6},-32768\leq a_i\leq 32767，m>0$。

Problem solving ideas:

First determine a state, dp[i][j] represents the maximum value of group i formed by the first j numbers, and the state transition equation is:

$$dp[i][j] = max(dp[i][j-1], max(dp[i-1][k]))+a[j](1\leq k < j)$$ The first part is to add the jth to the i-1th segment, and the latter part is that the ith segment starts with the jth element. But what makes this question a bit uncomfortable is that his m doesn't tell you the specific range, but this question can only be used $O(n*m)$ The complexity of the state transition equation dp[i][j] can be transferred from dp[i][j-1], it is only related to the previous state, so it can be replaced by a one-dimensional array, that is, $$dp[j] = max(dp[j-1],Max[j])+a[j](其中Max[j]代表dp[k](1\leq k<j)的最大值)$$ Then use Maxm to record the maximum value when it is divided into m segments, which is the final answer.

AC code:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL INF = 1LL << 50;
const int maxn = 1e6;
LL Max[maxn + 5];
LL dp[maxn + 5];
LL a[maxn + 5];
LL Maxm;
int n, m;
int main() {
    while(scanf("%d%d", &m, &n) != EOF) {
        for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);
        memset(dp, 0, sizeof(dp));
        memset(Max, 0, sizeof(Max));
        for(int i = 1; i <= m; i++) {
            Maxm = -INF;//Maxm每次初始化为-oo
            for(int j = i; j <= n; j++) {
                dp[j] = max(dp[j - 1], Max[j]) + a[j];
                Max[j] = Maxm;//Max[j]记录上一状态前j-1个的最大值。
                Maxm = max(Maxm, dp[j]);
            }
        }
        printf("%lld\n", Maxm);
    }
    return 0;
}