SPOJ GSS3 (动态dp)

题意

题目链接

Sol

这题可以动态dp做。

设\(f[i]\)表示以\(i\)为结尾的最大子段和，\(g[i]\)表示\(1-i\)的最大子段和

那么

\(f[i] = max(f[i - 1] + a[i], a[i])\)

\(g[i] = max(g[i - 1], f[i])\)

发现只跟前一项有关，而且\(g[i]从\)f[i]$转移过来的那一项可以直接拆开

那么构造矩阵

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

直接转移就行了

复杂度\(O(nlogn * 27)\)

#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\n", max(ans.m[2][1], ans.m[2][3]));
        }
    }
    return 0;
}
/*
4
-1 -2 -3 -4
2
1 1 4
1 1 2
*/

原文地址：https://www.cnblogs.com/zwfymqz/p/10425320.html