AT_abc438_d - Tail of Snake（DP）

题目链接：D - Tail of Snake

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题目大意：

从三个长度为n的序列A、B、C中选取分界点 x 和 y，

要求的和为 A数组的1~x，B数组的x+1~y，C数组的y+1~n，各段对应的和

求：各段对应序列和的最大值。

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思路：

跟爬山是同一类DP，

可以把A，B，C，从高到底看成三个阶梯，

只能往下走不能往上走

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代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#include<array>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/numeric>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define mst(a,x) memset(a,x,sizeof (a))
#define gmap __gnu_pbds::cc_hash_table
#define power __gnu_cxx::power
using namespace std;
typedef pair<int, int> pii;

const int N = 1000086, mod = 998244353;

int n, m;
int a[N], b[N], c[N];

void solve() {

    cin >> n;
    vector<array<int, 3>> f(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i];
    for (int i = 1; i <= n; i++) cin >> c[i];
    
    for (int i = 1; i <= n; i++) {
        f[i][0] = f[i - 1][0] + a[i];
        if (i >= 2) f[i][1] = max(f[i - 1][0], f[i - 1][1]) + b[i];
        if (i >= 3) f[i][2] = max(f[i - 1][1], f[i - 1][2]) + c[i];
    }
    
    cout << f[n][2] << endl;
=
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
    //  cin >> T;
    while (T--) solve();
    
    return 0;
}