洛谷 P1725 简单DP单调队列优化

题意：起点为1，终点为n，每个点都有一个权值ai，每次只能从i走到[i+l, i+r]，求能获得的最大权值

n <= 2e5

简单的dp，转移为dp[i] = max(dp[j]) + a[i]，但是显然直接暴力会T

倒推，单调队列维护一下跑过的dp[i]，更新前面的值

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<deque>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

const int mx = 2e5+10;
struct node{
    int a, id;
    node(int a = 0, int id = 0): a(a), id(id){}
};
deque<node> q;
int dp[mx], a[mx];

int main(){
    int n, l, r;
    scanf("%d%d%d", &n, &l, &r);
    for (int i = 0; i <= n; i++) scanf("%d", &a[i]);
    for (int i = n-l+1; i <= n; i++) dp[i] = a[i];
    for (int i = n+1; i >= l; i--){
        while (!q.empty() && dp[i] > q.back().a) q.pop_back();
        q.push_back(node(dp[i], i));
        dp[i-l] = q.front().a+a[i-l];
        if (i+r == q.front().id) q.pop_front();
    }
    printf("%d
", dp[0]);
    return 0;
}