uva11992 - Fast Matrix Operations 区间修改 二维线段树

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

  这个题真好。。

  任意区间都能分解成不超过2h个不相交区间的并，这是区间修改和点修改的区别。

  二维线段树，其实就是一些一维线段树。。每一行看成一个线段树，矩阵修改就变成了把每一行的线段树修改。这里add操作其实是不需要pushdown的，但是为了方便，少写点代码，不管是set还是add都pushdown。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define INF 0x3f3f3f3f
#define MAXN 25
#define MAXM 510
#define MAXNODE 1<<17
#define eps 1e-9
#define pi 4*atan(1.0)
#define pii pair<int,int>
using namespace std;
int T,R,C,M,op,x1,y1,x2,y2,v,_sum,_min,_max;

//修改、查询区间均为[y1,y2]
struct IntervalTree{
    //同一结点先考虑set，再考虑add
    //sumv[o],minv[o],maxv[o]保存的是以o为根结点的树内修改后的和，最大，最小值，o的父节点及以上结点修改的情况没有包括在里面
    int setv[MAXNODE],addv[MAXNODE],sumv[MAXNODE],minv[MAXNODE],maxv[MAXNODE];

    void maintain(int o,int L,int R){
        int lc=(o<<1),rc=(o<<1)+1;
        if(L<R){
            sumv[o]=sumv[lc]+sumv[rc];
            minv[o]=min(minv[lc],minv[rc]);
            maxv[o]=max(maxv[lc],maxv[rc]);
        }
        if(setv[o]>=0){
            sumv[o]=setv[o]*(R-L+1);
            minv[o]=maxv[o]=setv[o];
        }
        if(addv[o]){
            sumv[o]+=addv[o]*(R-L+1);
            minv[o]+=addv[o];
            maxv[o]+=addv[o];
        }
    }

    //把o结点的set和add都往下推一层
    void pushdown(int o){
        int lc=(o<<1),rc=(o<<1)+1;
        if(setv[o]>=0){
            setv[lc]=setv[rc]=setv[o];
            setv[o]=-1;
            addv[lc]=addv[rc]=0;
        }
        if(addv[o]){
            addv[lc]+=addv[o];
            addv[rc]+=addv[o];
            addv[o]=0;
        }
    }

    void update(int o,int L,int R){
        if(y1<=L&&y2>=R){
            if(op==1) addv[o]+=v;
            else{
                setv[o]=v;
                addv[o]=0;
            }
        }
        else{
            pushdown(o);
            int mid=(L+R)/2,lc=(o<<1),rc=(o<<1)+1;
            if(y1<=mid) update(lc,L,mid);
            else maintain(lc,L,mid);    //更新过的子树会自动maintain，没有更新的子树也要maintain，因为在pushdown的时候可能被修改
            if(y2>mid) update(rc,mid+1,R);
            else maintain(rc,mid+1,R);
        }
        maintain(o,L,R);
    }

    void query(int o,int L,int R,int add){    //add是o结点的父结点及以上结点增加的值，也要被加入o
        if(setv[o]>=0){    //set是个例外，不需要[L,R]包含在[y1,y2]中
            int v=setv[o]+addv[o]+add;
            _sum+=v*(min(R,y2)-max(L,y1)+1);    //两个区间的交集
            _min=min(_min,v);
            _max=max(_max,v);
        }
        else if(y1<=L&&y2>=R){
            _sum+=sumv[o]+add*(R-L+1);
            _min=min(_min,minv[o]+add);
            _max=max(_max,maxv[o]+add);
        }
        else{
            int mid=(L+R)/2,lc=(o<<1),rc=(o<<1)+1;
            if(y1<=mid) query(lc,L,mid,add+addv[o]);
            if(y2>mid) query(rc,mid+1,R,add+addv[o]);
        }
    }

}tree[MAXN];

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&R,&C,&M)!=EOF){
        memset(tree,0,sizeof(tree));
        for(int i=1;i<=R;i++){
            memset(tree[i].setv,-1,sizeof(tree[i].setv));
            tree[i].setv[1]=0;
        }
        while(M--){
            scanf("%d%d%d%d%d",&op,&x1,&y1,&x2,&y2);
            if(op<3){
                scanf("%d",&v);
                for(int i=x1;i<=x2;i++) tree[i].update(1,1,C);
            }
            else{
                _sum=0;
                _min=INF;
                _max=-INF;
                for(int i=x1;i<=x2;i++) tree[i].query(1,1,C,0);
                printf("%d %d %d\n",_sum,_min,_max);
            }
        }
    }
    return 0;
}