UVA 11992 二维线段树

There is a matrix containing at most 10⁶ elements divided into r rows and c columns. Each element has a location (x,y) where 1<=x<=r,1<=y<=c. Initially, all the elements are zero. You need to handle four kinds of operations:

1 x1 y1 x2 y2 v

Increment each element (x,y) in submatrix (x1,y1,x2,y2) by v (v>0)

2 x1 y1 x2 y2 v

Set each element (x,y) in submatrix (x1,y1,x2,y2) to v

3 x1 y1 x2 y2

Output the summation, min value and max value of submatrix (x1,y1,x2,y2)

In the above descriptions, submatrix (x1,y1,x2,y2) means all the elements (x,y) satisfying x1<=x<=x2 and y1<=x<=y2. It is guaranteed that 1<=x1<=x2<=r, 1<=y1<=y2<=c. After any operation, the sum of all the elements in the matrix does not exceed 10⁹.

Input

There are several test cases. The first line of each case contains three positive integers r, c, m, where m (1<=m<=20,000) is the number of operations. Each of the next m lines contains a query. There will be at most twenty rows in the matrix. The input is terminated by end-of-file (EOF). The size of input file does not exceed 500KB.

Output

For each type-3 query, print the summation, min and max.

Sample Input

4 4 8
1 1 2 4 4 5
3 2 1 4 4
1 1 1 3 4 2
3 1 2 4 4
3 1 1 3 4
2 2 1 4 4 2
3 1 2 4 4
1 1 1 4 3 3

Output for the Sample Input

45 0 5
78 5 7
69 2 7
39 2 7

题意：有一个r行c列的全0矩阵，支持一下3种操作

1 x1 y1 x2 y2 v 子矩阵所有元素增加v

2 x1 y1 x2 y2 v 子矩阵所有元素置v

3 x1 y1 x2 y2 v 查询子矩阵的元素和，最小值和最大值

思路：线段树懒惰标记的典型应用了，此题主要是练下二维线段树转换为一维的技巧、   

      若c列，则每行需要占用数组4*c。将每行都建一棵线段树即可。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1000800
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define inf 0x3f3f3f3f
int r,c,m;
struct ST
{
	int l,r,add,set,sum,maxkey,minkey;
}st[maxn<<2];

inline int max(int a,int b)
{
	return a>b?a:b;
}

inline int min(int a,int b)
{
	return a>b?b:a;
}

void PushUp(int k,int id)
{
	int t = 4*k*c;
	st[t+id].sum = st[t+(id<<1)].sum + st[t+(id<<1|1)].sum;
	st[t+id].maxkey = max(st[t+(id<<1)].maxkey,st[t+(id<<1|1)].maxkey);
	st[t+id].minkey = min(st[t+(id<<1)].minkey,st[t+(id<<1|1)].minkey);
}

void PushDown(int k,int id)
{
	int t = 4*k*c;
	if(st[t+id].set != -1)
	{
		st[t+(id<<1)].add = st[t+(id<<1|1)].add = 0;
		st[t+(id<<1)].set = st[t+(id<<1|1)].set = st[t+id].set;
		st[t+(id<<1)].maxkey = st[t+(id<<1)].minkey = st[t+id].set;
		st[t+(id<<1|1)].maxkey = st[t+(id<<1|1)].minkey = st[t+id].set;
		st[t+(id<<1)].sum = (st[t+(id<<1)].r - st[t+(id<<1)].l + 1) * st[t+id].set;
		st[t+(id<<1|1)].sum = (st[t+(id<<1|1)].r - st[t+(id<<1|1)].l + 1) * st[t+id].set;
		st[t+id].set = -1;
	}

	if(st[t+id].add)
	{
		st[t+(id<<1)].add += st[t+id].add;
		st[t+(id<<1|1)].add += st[t+id].add;
		st[t+(id<<1)].maxkey += st[t+id].add;
		st[t+(id<<1)].minkey += st[t+id].add;
		st[t+(id<<1)].sum += (st[t+(id<<1)].r - st[t+(id<<1)].l + 1) * st[t+id].add;
		st[t+(id<<1|1)].maxkey += st[t+id].add;
		st[t+(id<<1|1)].minkey += st[t+id].add;
		st[t+(id<<1|1)].sum += (st[t+(id<<1|1)].r - st[t+(id<<1|1)].l + 1) * st[t+id].add;
		st[t+id].add = 0;
	}
}

void buildtree(int k,int id,int l,int r)
{
	int t = 4*k*c;
	st[t+id].l = l,st[t+id].r = r;
	st[t+id].set = -1;st[t+id].add = 0;
	st[t+id].sum = st[t+id].maxkey = st[t+id].minkey = 0;
	if(l == r)	return;
	int mid = (l+r) >> 1;
	buildtree(k,lson);
	buildtree(k,rson);
	PushUp(k,id);
}

void Update(int k,int id,int l,int r,int key,int ope)
{
	int t = 4*k*c;
	if(st[t+id].l == l && st[t+id].r == r)
	{
		if(ope == 1)//ADD
		{
			st[t+id].add += key;
			st[t+id].maxkey += key;
			st[t+id].minkey += key;
			st[t+id].sum += (r - l + 1) * key;
		}
		else//SET
		{
			st[t+id].add = 0;
			st[t+id].set = key;
			st[t+id].maxkey = st[t+id].minkey = key;
			st[t+id].sum = (r - l + 1) * key;
		}
		return;
	}
	PushDown(k,id);
	if(st[t+(id<<1)].r >= r)
	{
		Update(k,id<<1,l,r,key,ope);
	}
	else if(st[t+(id<<1|1)].l <= l)
	{
		Update(k,id<<1|1,l,r,key,ope);
	}
	else 
	{
		Update(k,id<<1,l,st[t+(id<<1)].r,key,ope);
		Update(k,id<<1|1,st[t+(id<<1|1)].l,r,key,ope);
	}
	PushUp(k,id);
}

int Query(int k,int id,int l,int r,int ope)
{
	int t = 4*k*c;
	if(st[t+id].l == l && st[t+id].r == r)
	{
		switch(ope)
		{
		case(1):return st[t+id].sum;
		case(2):return st[t+id].minkey;
		case(3):return st[t+id].maxkey;
		}
	}
	PushDown(k,id);
	if(st[t+(id<<1)].r >= r)
		return Query(k,id<<1,l,r,ope);
	else if(st[t+(id<<1|1)].l <= l)
		return Query(k,id<<1|1,l,r,ope);
	else 
	{
		switch(ope)
		{
		case(1):return Query(k,id<<1,l,st[t+(id<<1)].r,ope) + Query(k,id<<1|1,st[t+(id<<1|1)].l,r,ope);
		case(2):return min(Query(k,id<<1,l,st[t+(id<<1)].r,ope),Query(k,id<<1|1,st[t+(id<<1|1)].l,r,ope));
		case(3):return max(Query(k,id<<1,l,st[t+(id<<1)].r,ope),Query(k,id<<1|1,st[t+(id<<1|1)].l,r,ope));
		}
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	int x1,x2,y1,y2,v,ope;
	while(scanf("%d%d%d",&r,&c,&m)==3)
	{
		for(int i = 1;i <= r;i++)	buildtree(i-1,1,1,c);
		while(m--)
		{
			scanf("%d%d%d%d%d",&ope,&x1,&y1,&x2,&y2);
			if(ope != 3)
			{
				scanf("%d",&v);
				for(int i = x1;i <= x2;i++)	Update(i-1,1,y1,y2,v,ope);
			}
			else 
			{
				int ans = 0,Min = inf,Max = -inf;
				for(int i = x1;i <= x2;i++)
				{
					ans += Query(i-1,1,y1,y2,1);
					Min = min(Min,Query(i-1,1,y1,y2,2));
					Max = max(Max,Query(i-1,1,y1,y2,3));
				}
				printf("%d %d %d\n",ans,Min,Max);
			}
		}
	}
	return 0;
}