[SPOJ 1716] Can you answer these queries III（动态 DP） | 错题本

题目

Can you answer these queries III

分析

本题当然可以用传统线段树做
记 $f_i$ 表示以 $a_i$ 结尾的最大子段和， $g_i$ 表示前 $i$ 个数的最大子段和。则 $\begin{aligned} f_i &= \max\{f_{i - 1} + a_i, a_i\} \\ g_i &= \max\{f_i, g_{i - 1}\} \\ &=\max\{f_{i - 1} + a_i, a_i, g_{i - 1}\} \end{aligned}$ 更改一下矩阵乘法的定义（ $+ \to \max, \times \to +$）不难得到矩阵转移（我习惯列乘行） $\begin{pmatrix} f_i \\ g_i \\ 0 \end{pmatrix} = \begin{pmatrix} f_{i - 1} \\ g_{i - 1} \\ 0 \end{pmatrix} \times \begin{pmatrix} a_i & -\infty & a_i \\ a_i & 0 & a_i \\ -\infty & -\infty & 0 \end{pmatrix}$
线段树维护区间矩阵乘积即可。

错因

• 算矩阵乘法时初始矩阵应初始化为全 $-\infty$，而不是全 $0$。

代码

注意矩阵初始化和计算答案的细节。

#include <algorithm>
#include <cstdio>
#include <cstring>

const int INF = 0x3f3f3f3f;
const int MAXN = 50000;
const int MAXP = 3;

int N, Q;
int A[MAXN + 5];

struct Matrix {
	int n, m;
	int Mat[MAXP + 5][MAXP + 5];

	Matrix(int _n = 0, int _m = 0) {
		n = _n, m = _m;
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				Mat[i][j] = -INF;
	}

	int* operator [] (const int &i) {
		return Mat[i];
	}

	Matrix operator * (Matrix other) {
		Matrix ret(n, other.m);
		for (int i = 1; i <= ret.n; i++)
			for (int j = 1; j <= ret.m; j++)
				for (int k = 1; k <= m; k++)
					ret[i][j] = std::max(ret[i][j], Mat[i][k] + other[k][j]);
		return ret;
	}

	void Debug() {
		for (int i = 1; i <= n; i++, puts(""))
			for (int j = 1; j <= m; j++)
				printf("%d ", Mat[i][j]);
	}
};

struct SegmentTree {
	#define lch (u << 1)
	#define rch (u << 1 | 1)

	Matrix Dp[(MAXN << 2) + 5];

	void PushUp(int u) {
		Dp[u] = Dp[lch] * Dp[rch];
	}

	void Build(int u, int lft, int rgt) {
		if (lft == rgt) {
			Dp[u].n = Dp[u].m = 3;
			Dp[u][2][2] = Dp[u][3][3] = 0;
			Dp[u][1][1] = Dp[u][1][3] = A[lft];
			Dp[u][2][1] = Dp[u][2][3] = A[lft];
			Dp[u][1][2] = Dp[u][3][1] = Dp[u][3][2] = -INF;
			return;
		}
		int mid = (lft + rgt) >> 1;
		Build(lch, lft, mid);
		Build(rch, mid + 1, rgt);
		PushUp(u);
	}

	void Modify(int u, int lft, int rgt, int pos, int val) {
		if (lft == rgt) {
			Dp[u][1][1] = Dp[u][1][3] = val;
			Dp[u][2][1] = Dp[u][2][3] = val;
			return;
		}
		int mid = (lft + rgt) >> 1;
		if (pos <= mid)
			Modify(lch, lft, mid, pos, val);
		else
			Modify(rch, mid + 1, rgt, pos, val);
		PushUp(u);
	}

	Matrix Query(int u, int lft, int rgt, int l, int r) {
		if (l <= lft && rgt <= r)
			return Dp[u];
		int mid = (lft + rgt) >> 1;
		if (mid < l)
			return Query(rch, mid + 1, rgt, l, r);
		if (r <= mid)
			return Query(lch, lft, mid, l, r);
		return Query(lch, lft, mid, l, r) * Query(rch, mid + 1, rgt, l, r);
	}

	#undef lch
	#undef rch
}T;

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; i++)
		scanf("%d", &A[i]);
	T.Build(1, 1, N);
	scanf("%d", &Q);
	while (Q--) {
		int opt, x, y;
		scanf("%d%d%d", &opt, &x, &y);
		if (opt == 0)
			T.Modify(1, 1, N, x, y);
		else {
			Matrix Ans = T.Query(1, 1, N, x, y);
			printf("%d\n", std::max(Ans[2][1], Ans[2][3]));
		}
	}
	return 0;
}