【SP1716】GSS3 - Can you answer these queries III（动态DP）

题目链接

之前用线段树写了一遍，现在用\(ddp\)再写一遍。

#include <cstdio>
#define lc (now << 1)
#define rc (now << 1 | 1)
inline int max(int a, int b){
	return a > b ? a : b;
}
const int INF = 2147483647 >> 2;
const int MAXN = 50010;
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
struct Matrix{
	int a[3][3];
}sum[MAXN << 2];
int a[MAXN], n, m, A, B, opt;
inline Matrix operator * (Matrix a, Matrix b){
	Matrix c;
	for(int i = 0; i < 3; ++i)
		for(int j = 0; j < 3; ++j)
			c.a[i][j] = -INF;
	for(int i = 0; i < 3; ++i)
		for(int j = 0; j < 3; ++j)
			for(int k = 0; k < 3; ++k)
				c.a[i][j] = max(c.a[i][j], a.a[i][k] + b.a[k][j]);
	return c;
}
inline void pushup(int now){
	sum[now] = sum[lc] * sum[rc];
}
void build(int now, int l, int r){
	if(l == r){
		sum[now].a[0][0] = sum[now].a[0][2] = sum[now].a[1][0] = sum[now].a[1][2] = a[l];
		sum[now].a[0][1] = sum[now].a[2][0] = sum[now].a[2][1] = -INF;
		sum[now].a[1][1] = sum[now].a[2][2] = 0; return ;
	}
	int mid = (l + r) >> 1;
	build(lc, l, mid); build(rc, mid + 1, r); pushup(now);
}
void modify(int now, int l, int r, int x, int y){
	if(l == r){
		sum[now].a[0][0] = sum[now].a[0][2] = sum[now].a[1][0] = sum[now].a[1][2] = y;
		return ;
	}
	int mid = (l + r) >> 1;
	if(x <= mid) modify(lc, l, mid, x, y);
	else modify(rc, mid + 1, r, x, y);
	pushup(now);
}
Matrix query(int now, int l, int r, int wl, int wr){
	if(l >= wl && r <= wr) return sum[now];
	int mid = (l + r) >> 1;
	if(wl <= mid && wr > mid)
		 return query(lc, l, mid, wl, wr) * query(rc, mid + 1, r, wl ,wr);
	else if(wl <= mid)
		 return query(lc, l, mid, wl, wr);
	else return query(rc, mid + 1, r, wl, wr);
}
int main(){
	n = read();
	for(int i = 1; i <= n; ++i)
		a[i] = read();
	build(1, 1, n);
	m = read();
	for(int i = 1; i <= m; ++i){
		opt = read(); A = read(); B = read();
		if(opt){
			Matrix ans = query(1, 1, n, A, B);
			printf("%d\n", max(ans.a[1][0], ans.a[1][2]));
		}
		else modify(1, 1, n, A, B);
	}
	return 0;
}