UVA 11992 Fast Matrix Operations

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本文介绍了一种高效的矩阵操作方法，支持矩阵元素的增量更新、赋值及子矩阵的求和、极值查询等操作，适用于处理大规模矩阵数据结构问题。

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

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!

题目大意：有一个r行c列的全0矩阵，支持三种操作：

1 x1 y1 x2 y2 v 子矩阵(x1，y1，x2，y2)的所有元素增加v（v > 0）。

2 x1 y1 x2 y2 v 子矩阵(x1，y1，x2，y2)的所有元素设为v（v > 0）。

3 x1 y1 x2 y2 查询子矩阵(x1，y1，x2，y2)的元素和、最小值和最大值。

子矩阵(x1，y1，x2，y2)是指满足x1 <= x <= x2，y1 <= y <= y2，的所有元素(x，y)。输入保证任何时刻矩阵内的所有元素之和不超过10 ^ 9。

其中，矩阵不超过20行，元素总数不超过10 ^ 6，且操作数不超过20000。

每一行建一个线段树有意思

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAX_N = 100010;
const int INF = 0x3f3f3f3f;
int s[30][MAX_N<<2],MAX[30][MAX_N<<2],MIN[30][MAX_N<<2],addv[30][MAX_N<<2],setv[30][MAX_N<<2];
void up(int i,int p){
        s[i][p]=s[i][p*2]+s[i][p*2+1];
        MIN[i][p]=min(MIN[i][p*2],MIN[i][p*2+1]);
        MAX[i][p]=max(MAX[i][p*2],MAX[i][p*2+1]);
}
void down(int i,int p,int l,int r){
    int mid = (l+r)>>1;
    if(setv[i][p]){
        setv[i][p*2] = setv[i][p];
        setv[i][p*2+1] = setv[i][p];
        addv[i][p*2]=addv[i][p*2+1]=0;
        s[i][p*2]=setv[i][p*2]*(mid-l+1);
        s[i][p*2+1]=setv[i][p*2+1]*(r-mid);
        MIN[i][p*2]=MIN[i][p*2+1]=setv[i][p];
        MAX[i][p*2]= MAX[i][p*2+1]=setv[i][p];
        setv[i][p] = 0;
    }
    if(addv[i][p]){
        addv[i][p*2] += addv[i][p];
        addv[i][p*2+1] += addv[i][p];
        s[i][p*2]+=addv[i][p]*(mid-l+1);
        s[i][p*2+1]+=addv[i][p]*(r-mid);
        MIN[i][p*2]+=addv[i][p*2];
        MIN[i][p*2+1]+=addv[i][p*2+1];
        MAX[i][p*2]+=addv[i][p];
        MAX[i][p*2+1]+=addv[i][p];
        addv[i][p] = 0;
    }
}
void build(int i,int p,int l,int r){
    addv[i][p]=0;
    setv[i][p]=0;
    if(l==r){
        MIN[i][p]=MAX[i][p]=s[i][p]=0;
        return;
    }
    int mid =(l+r)>>1;
    build(i,p*2,l,mid);
    build(i,p*2+1,mid+1,r);
    up(i,p);
}
void change(int a,int i,int p,int l,int r,int x,int y,int v){
    if(x<=l&&r<=y){
        if(a==1){
            addv[i][p]+=v;
            s[i][p]+=v*(r-l+1);
            MIN[i][p]+=v;
            MAX[i][p]+=v;
        }
        else if(a==2){
            addv[i][p]=0;
            setv[i][p]=v;
            s[i][p]=v*(r-l+1);
            MIN[i][p]=MAX[i][p]=v;
        }
        return;
    }
    down(i,p,l,r);
    int mid=(l+r)>>1;
    if(x<=mid) change(a,i,p*2,l,mid,x,y,v);
    if(y>mid) change(a,i,p*2+1,mid+1,r,x,y,v);
    up(i,p);
}
void query(int i,int p,int l,int r,int x,int y,int& sum,int& maxx,int& minn){
  if(x<=l&&r<=y){
    sum=s[i][p];
    maxx=MAX[i][p];
    minn=MIN[i][p];
    return;
  }
down(i,p,l,r);
int mid=(l+r)>>1;
int ts,tmax,tmin;
  sum=0;maxx=-1;minn=INF;
  if(x<=mid) {
    query(i,p*2,l,mid,x,y,ts,tmax,tmin);
    sum+=ts;
    maxx=max(maxx,tmax);
    minn=min(minn,tmin);
  }
  if(y>mid){
    query(i,p*2+1,mid+1,r,x,y,ts,tmax,tmin);
    sum+=ts;
    maxx=max(maxx,tmax);
    minn=min(minn,tmin);
  }
}


int main(){
    int r,c,m;
    while(~scanf("%d%d%d",&r,&c,&m)){
        for(int i=i;i<=r;++i)
           build(i,1,1,c);
    while(m--){
        int a,x1,y1,x2,y2,v;
        scanf("%d",&a);
        if(a==1||a==2){
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
            for(int i=x1;i<=x2;++i)
               change(a,i,1,1,c,y1,y2,v);
        }
        else if(a==3){
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
           int sum = 0,minn=INF,maxx=-1;
           int ts,tmin,tmax;
           for(int i=x1;i<=x2;i++){
            query(i,1,1,c,y1,y2,ts,tmax,tmin);
            sum+=ts;
            maxx=max(maxx,tmax);
            minn=min(minn,tmin);
           }
           printf("%d %d %d\n",sum,minn,maxx);
        }
    }
    }
    return 0;
}