Problem F
Fast Matrix Operations
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
Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yeji Shen, Dun Liang
Note: Please make sure to test your program with the gift I/O files before submitting!
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模板
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#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 555555
using namespace std;
const int OO=1e9;
int num[30][N];
int _min,_max,_sum;
struct Tree
{
int l;
int r;
int max;
int min;
int sum;
int add;
int set;
}big_tree[21][N*4];
void push_up(int root,Tree tree[])
{
tree[root].max=max(tree[root<<1].max,tree[root<<1|1].max);
tree[root].min=min(tree[root<<1].min,tree[root<<1|1].min);
tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;
}
void push_down(int root,Tree tree[])
{
if (tree[root].set!=-1)
{
if (tree[root].l!=tree[root].r)
{
//传递懒惰标记
tree[root<<1].add=tree[root<<1|1].add=0;
tree[root<<1].set=tree[root<<1|1].set=tree[root].set;
//最更新大值
tree[root<<1].max=tree[root<<1|1].max=tree[root].set;
//更新最小值
tree[root<<1].min=tree[root<<1|1].min=tree[root].set;
//更新区间和
tree[root<<1].sum=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].set;
tree[root<<1|1].sum=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].set;
}
tree[root].set=-1;
}
if (tree[root].add>0)
{
if (tree[root].l!=tree[root].r)
{
//传递懒惰标记
tree[root<<1].add+=tree[root].add;
tree[root<<1|1].add+=tree[root].add;
//更新最大值
tree[root<<1].max+=tree[root].add;
tree[root<<1|1].max+=tree[root].add;
//更新最小值
tree[root<<1].min+=tree[root].add;
tree[root<<1|1].min+=tree[root].add;
//更新区间和
tree[root<<1].sum+=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].add;
tree[root<<1|1].sum+=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].add;
}
tree[root].add=0;
}
}
void build(int root,int l,int r,Tree tree[])
{
tree[root].l=l;
tree[root].r=r;
if(tree[root].l==tree[root].r)
{
tree[root].max=0;
tree[root].min=0;
tree[root].sum=0;
tree[root].add=0;
tree[root].set=-1;
return;
}
int mid=(l+r)/2;
build(root<<1,l,mid,tree);
build(root<<1|1,mid+1,r,tree);
push_up(root,tree);
}
void update_add(int root,int L,int R,int val,Tree tree[])
{
if(L<=tree[root].l&&R>=tree[root].r)
{
tree[root].add+=val;
tree[root].max+=val;
tree[root].min+=val;
tree[root].sum+=(tree[root].r-tree[root].l+1)*val;
return;
}
push_down(root,tree);
int mid=(tree[root].l+tree[root].r)/2;
if(L<=mid)
update_add(root<<1,L,R,val,tree);
if (R>mid)
update_add(root<<1|1,L,R,val,tree);
push_up(root,tree);
}
void update_set(int root,int L,int R,int val,Tree tree[])
{
if(L<=tree[root].l&&R>=tree[root].r)
{
tree[root].set=val;
tree[root].add=0;
tree[root].max=val;
tree[root].min=val;
tree[root].sum=(tree[root].r-tree[root].l+1)*val;
return;
}
push_down(root,tree);
int mid=(tree[root].l+tree[root].r)/2;
if(L<=mid)
update_set(root<<1,L,R,val,tree);
if (R>mid)
update_set(root<<1|1,L,R,val,tree);
push_up(root,tree);
}
void query(int root,int L,int R,Tree tree[])
{
if(L<=tree[root].l&&R>=tree[root].r)
{
_min=min(_min,tree[root].min);
_max=max(_max,tree[root].max);
_sum+=tree[root].sum;
return;
}
push_down(root,tree);
int mid=(tree[root].l+tree[root].r)/2;
if(L<=mid) query(root<<1,L,R,tree);
if(R>mid) query(root<<1|1,L,R,tree);
}
int main()
{
int r,c,m;
while (~scanf("%d%d%d",&r,&c,&m))
{
int tp,x1,x2,y1,y2;
for (int i=1;i<=r;i++)
{
build(1,1,c,big_tree[i]);
}
while (m--)
{
int v;
scanf("%d%d%d%d%d",&tp,&x1,&y1,&x2,&y2);
if (tp==1)
{
scanf("%d",&v);
for (int i=x1;i<=x2;i++)
{
update_add(1,y1,y2,v,big_tree[i]);
}
}
if (tp==2)
{
scanf("%d",&v);
for (int i=x1;i<=x2;i++)
{
update_set(1,y1,y2,v,big_tree[i]);
}
}
if (tp==3)
{
_max=0;
_min=OO;
_sum=0;
for (int i=x1;i<=x2;i++)
{
query(1,y1,y2,big_tree[i]);
}
printf("%d %d %d\n",_sum,_min,_max);
}
}
}
return 0;
}