BNU20410 UVA11992 线段树区间更新

Fast Matrix Operations

Time Limit: 5000ms

Memory Limit: 131072KB
This problem will be judged on UVA.
Original ID: 11992

64-bit integer IO format: %lld     
Java class name: Main

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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

https://www.bnuoj.com/v3/problem_show.php?pid=20410

给你一个初始全是0的矩阵r*c    三个操作1把满足 x1<=x<=x2  y1<=y<y2    (x,y)的子矩阵都加V 

2就把子矩阵变成V  3就是查询子矩阵和,最小值,最大值;

多维转换成一维  注意set的操作放下去的时候add标记得清零

#include<stdio.h>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int SUM,MIN,MAX;
struct node
{
int minn,maxn,sum,add,st;
} tree[1000010*4];

void pushup(int rt)
{
int lc=rt<<1,rc=rt<<1|1;
tree[rt].sum=tree[lc].sum+tree[rc].sum;
tree[rt].minn=min(tree[lc].minn,tree[rc].minn);
tree[rt].maxn=max(tree[lc].maxn,tree[rc].maxn);
}

void build(int l,int r,int rt)
{
tree[rt].add=0;
tree[rt].st=-1;
tree[rt].sum=0;
tree[rt].minn=0;
tree[rt].maxn=0;
if(l==r) return ;
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}

void pushdown(int rt,int len)
{
int lc=(rt<<1);
int rc=(rt<<1|1);
if(tree[rt].st>=0)
{
tree[lc].st=tree[rc].st=tree[rt].st;
tree[lc].add=tree[rc].add=0;//add清除
tree[rt].st=-1;
tree[lc].sum=tree[lc].st*(len-(len>>1));
tree[rc].sum=tree[rc].st*(len>>1);
tree[lc].minn=tree[lc].st;
tree[lc].maxn=tree[lc].st;
tree[rc].minn=tree[rc].st;
tree[rc].maxn=tree[rc].st;
}
if(tree[rt].add)
{
tree[lc].add+=tree[rt].add;
tree[rc].add+=tree[rt].add;
tree[lc].sum+=tree[rt].add*(len-(len>>1));//放下去个一半
tree[rc].sum+=tree[rt].add*(len>>1);//只要加上上层传下来的
tree[lc].minn+=tree[rt].add;
tree[lc].maxn+=tree[rt].add;
tree[rc].minn+=tree[rt].add;
tree[rc].maxn+=tree[rt].add;
tree[rt].add=0;
}
}

void update1(int L,int R,int l,int r,int rt,int p)
{
if(L<=l&&R>=r)
{
tree[rt].add+=p; //本层更新
tree[rt].sum+=(r-l+1)*p;
tree[rt].minn+=p;
tree[rt].maxn+=p;
return ;
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) update1(L,R,lson,p);
if(R>m) update1(L,R,rson,p);
pushup(rt); //更新返回
}

void update2(int L,int R,int l,int r,int rt,int p)
{
if(L<=l&&R>=r)
{
tree[rt].add=0;
tree[rt].st=p;
tree[rt].sum=(r-l+1)*p;
tree[rt].minn=p;
tree[rt].maxn=p;
return ;
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) update2(L,R,lson,p);
if(R>m) update2(L,R,rson,p);
pushup(rt);
}

int query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
MIN=min(MIN,tree[rt].minn);
MAX=max(MAX,tree[rt].maxn);
return tree[rt].sum;
}
pushdown(rt,r-l+1);
int ret=0;
int m=(l+r)>>1;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson);
return ret;
}

int main()
{
int r,c,q,s,i,x1,x2,y1,y2,v,n;
while(scanf("%d%d%d",&r,&c,&q)!=EOF)
{
n=r*c;
build(1,n,1);
while(q--)
{
scanf("%d",&s);
if(s!=3)
{
if(s==1)
{
scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&v);
for(i=x1-1; i<x2; i++)
update1(i*c+y1,i*c+y2,1,n,1,v);
}
else
{
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
for(i=x1-1; i<x2; i++)
update2(i*c+y1,i*c+y2,1,n,1,v);
}
}
else
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
SUM=0;
MIN=0x7f7f7f7f;
MAX=0;
for(i=x1-1; i<x2; i++)
SUM+=query(i*c+y1,i*c+y2,1,n,1);
printf("%d %d %d\n",SUM,MIN,MAX);

}
}
}
return 0;
}