SPOJ GSS3 (動態dp)

自為風月馬前卒發表於2019-02-24
S3

題意

題目連結

Sol

這題可以動態dp做。

設(f[i])表示以(i)為結尾的最大子段和，(g[i])表示(1-i)的最大子段和

那麼

(f[i] = max(f[i – 1] + a[i], a[i]))

(g[i] = max(g[i – 1], f[i]))

發現只跟前一項有關，而且(g[i]從)f[i]$轉移過來的那一項可以直接拆開

那麼構造矩陣

[
egin{bmatrix}
a_{i} & -infty & dots a_{i} \
a_{i}, & 0 & a_{i}\
-infty, & -infty & 0 \
end{bmatrix}
]

直接轉移就行了

複雜度(O(nlogn * 27))

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9;
template<typename A, typename B> inline bool chmax(A &x, B y) {return x < y ? x = y, 1 : 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
    return x * f;
}
struct Ma {
    int m[4][4];
    Ma() {
        memset(m, -0x3f, sizeof(m));
    }
    Ma operator * (const Ma &rhs) const {
        Ma ans;
        for(int i = 1; i <= 3; i++)
            for(int j = 1; j <= 3; j++)
                for(int k = 1; k <= 3; k++)
                    chmax(ans.m[i][j], m[i][k] + rhs.m[k][j]);
        return ans; 
    }
    void init(int v) {
        m[1][1] = v; m[1][2] = -INF; m[1][3] = v;
        m[2][1] = v; m[2][2] = 0;    m[2][3] = v;
        m[3][1] = -INF; m[3][2] = -INF; m[3][3] = 0;
    }
}m[MAXN];
int N, M, a[MAXN];
#define ls k << 1
#define rs k << 1 | 1
void update(int k) {
    m[k] = m[ls] * m[rs];
}
void Build(int k, int l, int r) {
    if(l == r) {m[k].init(a[l]); return ;}
    int mid = l + r >> 1;
    Build(ls, l, mid); Build(rs, mid + 1, r);
    update(k);
}
void Modify(int k, int l, int r, int p, int v) {
    if(l == r) {m[k].init(v); return ;}
    int mid = l + r >> 1;
    if(p <= mid) Modify(ls, l, mid, p, v);
    else Modify(rs, mid + 1, r, p, v);
    update(k);
}
Ma Query(int k, int l, int r, int ql, int qr) {
    if(ql <= l && r <= qr) 
        return m[k];
    int mid = l + r >> 1;
    if(ql > mid) return Query(rs, mid + 1, r, ql, qr);
    else if(qr <= mid) return Query(ls, l, mid, ql, qr);
    else return (Query(ls, l, mid, ql, qr) * Query(rs, mid + 1, r, ql, qr));
}
int main() {
    N = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    Build(1, 1, N);
    M = read();
    while(M--) {
        int opt = read(), x = read(), y = read();
        if(opt == 0) Modify(1, 1, N, x, y);
        else {
            Ma ans = Query(1, 1, N, x, y);
            printf("%d
", max(ans.m[2][1], ans.m[2][3]));
        }
    }
    return 0;
}
/*
4
-1 -2 -3 -4
2
1 1 4
1 1 2
*/

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