线段树(多维+双成段更新) UVA 11992 Fast Matrix Operations

题目传送门

题意:训练指南P207

分析:因为矩阵不超过20行,所以可以建20条线段的线段树,支持两个区间更新以及区间查询.

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 20 + 5;
const int M = 5e4 + 5;
struct Segment_Tree	{
	struct Node	{
		int sum, mx, mn, add, setv;
	}node[M<<2];
	void _max(int &a, int b)	{
		if (a < b)	a = b;
	}
	void _min(int &a, int b)	{
		if (a > b)	a = b;
	}
	void push_up(int o)	{
		node[o].sum = node[o<<1].sum + node[o<<1|1].sum;
		node[o].mx = max (node[o<<1].mx, node[o<<1|1].mx);
		node[o].mn = min (node[o<<1].mn, node[o<<1|1].mn);
	}
	void push_down(int l, int r, int o)	{
		int len = r - l + 1;
		if (node[o].setv != -1 && l < r)	{					//先set,然后再add
			node[o<<1].setv = node[o<<1|1].setv = node[o].setv;
			node[o<<1].add = node[o<<1|1].add = 0;				//很重要的地方,有set的清除add,意思是set前的add都不要了
			node[o<<1].sum = node[o].setv * (len - (len >> 1));
			node[o<<1].mx = node[o<<1].mn = node[o].setv;
			node[o<<1|1].sum = node[o].setv * (len >> 1);
			node[o<<1|1].mx = node[o<<1|1].mn = node[o].setv;
			node[o].setv = -1;
		}
		if (node[o].add != 0 && l < r)	{
			node[o<<1].add += node[o].add;	node[o<<1|1].add += node[o].add;
			node[o<<1].sum += node[o].add * (len - (len >> 1));
			node[o<<1].mx += node[o].add;	node[o<<1].mn += node[o].add;
			node[o<<1|1].sum += node[o].add * (len >> 1);
			node[o<<1|1].mx += node[o].add;	node[o<<1|1].mn += node[o].add;
			node[o].add = 0;
		}
	}
	void build(int l, int r, int o)	{
		node[o].add = 0;	node[o].setv = -1;
		if (l == r)	{
			node[o].sum = 0;	node[o].mx = 0;	node[o].mn = 0;
			return ;
		}
		int mid = l + r >> 1;
		build (lson);	build (rson);
		push_up (o);
	}
	void updata(int ql, int qr, int op, int c, int l, int r, int o)	{
		if (ql <= l && r <= qr)	{
			if (op == 1)	{
				node[o].sum += c * (r - l + 1);	node[o].mx += c;	node[o].mn += c;
				node[o].add += c;	return ;
			}
			else if (op == 2)	{
				node[o].sum = c * (r - l + 1);	node[o].mx = node[o].mn = c;
				node[o].add = 0;
				node[o].setv = c;	return ;
			}
		}
		push_down (l, r, o);
		int mid = l + r >> 1;
		if (ql <= mid)	updata (ql, qr, op, c, lson);
		if (qr > mid)	updata (ql, qr, op, c, rson);
		push_up (o);
	}
	void query(int ql, int qr, int &_sum, int &_mn, int &_mx, int l, int r, int o)	{
		if (ql <= l && r <= qr)	{
			_sum += node[o].sum;
			_max (_mx, node[o].mx);
			_min (_mn, node[o].mn);
			return ;
		}
		push_down (l, r, o);
		int mid = l + r >> 1;
		if (ql <= mid)	query (ql, qr, _sum, _mn, _mx, lson);
		if (qr > mid)	query (ql, qr, _sum, _mn, _mx, rson);
	}
}st[N];
int n, m, q;

int main(void)	{
	while (scanf ("%d%d%d", &n, &m, &q) == 3)	{
		for (int i=1; i<=n; ++i)	{
			st[i].build (1, m, 1);
		}
		int x1, y1, x2, y2, v, op;
		while (q--)	{
			scanf ("%d%d%d%d%d", &op, &x1, &y1, &x2, &y2);
			if (op <= 2)	{
				scanf ("%d", &v);
				for (int i=x1; i<=x2; ++i)	{
					st[i].updata (y1, y2, op, v, 1, m, 1);
				}
			}
			else	{
				int sum = 0, mx = 0, mn = INF;
				for (int i=x1; i<=x2; ++i)	{
					st[i].query (y1, y2, sum, mn, mx, 1, m, 1);
				}
				printf ("%d %d %d\n", sum, mn, mx);
			}
		}
	}

	return 0;
}