Uva 11992 - Fast Matrix Operations题解(线段树区间更新+区间Set+Add,查询最大值,最小值,总和)

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 81 1 2 4 4 53 2 1 4 41 1 1 3 4 23 1 2 4 43 1 1 3 42 2 1 4 4 23 1 2 4 41 1 1 4 3 3

Output for the Sample Input

45 0 578 5 769 2 739 2 7

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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!

题意：

1 x1 y1 x2 y2 v 给矩阵 x1 y1 x2 y2的所有值加上v

2 x1 y1 x2 y2 v 把矩阵x1 y1 x2 y2置为(Set)v

3 x1 y1 x2 y2 查询矩阵x1 y1 x2 y2的总和,最小值，最大值

题解：

由于题目保证不超过20行，矩阵的数据总的不超过10^6个，有m=2*10^5次操作，那么用线段树做.

给每行建一棵线段树。

注意：

在add操作中不清除Set标记，在Set操作中要清除add标记。在query操作中要综合考虑add和Set。

在Set和add操作中，找到相应的区间后要先考虑Set，在考虑add(因为在Set的时候会把add的值清空，那么add是Set之后的操作)

时间复杂度

mlogn

#include<stdio.h>#include<algorithm>#include<math.h>#define MAX 1000000#define LL long long int#define INF 0x7f7f7fusing namespace std;int tMin,tMax;struct node{    int left,right;    int sum, add;    int Min,Max;    int Set;//Set=-1表示该段没有被重置}arr[21][3*MAX];void build(int idx, int l, int r,int pos){    arr[pos][idx].left = l, arr[pos][idx].right = r;    arr[pos][idx].add = 0;    arr[pos][idx].Min=0;    arr[pos][idx].Max=0;    arr[pos][idx].sum=0;    arr[pos][idx].Set=-1;    if(l == r){        return ;    }    int mid = (l + r) >> 1;    build(idx * 2, l, mid,pos);    build(idx * 2 + 1, mid + 1, r,pos);}void pushDown(int idx,int pos){    int lc=idx*2;    int rc=idx*2+1;    if(arr[pos][idx].Set>=0){        arr[pos][lc].Set=arr[pos][rc].Set=arr[pos][idx].Set;        arr[pos][lc].add=arr[pos][rc].add=0;        arr[pos][idx].Set=-1;        arr[pos][lc].sum=(arr[pos][lc].right - arr[pos][lc].left + 1)*arr[pos][lc].Set;//由于是Set那么sum的值要重新赋值        arr[pos][rc].sum=(arr[pos][rc].right - arr[pos][rc].left + 1)*arr[pos][rc].Set;        arr[pos][lc].Min=arr[pos][lc].Set;//由于是Set，那么就直接复制        arr[pos][lc].Max=arr[pos][lc].Set;        arr[pos][rc].Min=arr[pos][rc].Set;        arr[pos][rc].Max=arr[pos][rc].Set;    }    if(arr[pos][idx].add>0){//注意此处的处理，免得加了两边,一定要先计算完了，在吧arr[pos][idx].add=0        arr[pos][lc].add+=arr[pos][idx].add;        arr[pos][rc].add+=arr[pos][idx].add;        arr[pos][lc].sum += (arr[pos][lc].right - arr[pos][lc].left + 1) * arr[pos][idx].add;        arr[pos][rc].sum += (arr[pos][rc].right - arr[pos][rc].left + 1) * arr[pos][idx].add;        arr[pos][lc].Min+=arr[pos][idx].add;//由于整段都增加add，那么最大值和最小值加上add，即可        arr[pos][lc].Max+=arr[pos][idx].add;        arr[pos][rc].Min+=arr[pos][idx].add;        arr[pos][rc].Max+=arr[pos][idx].add;        arr[pos][idx].add=0;    }}void set(int idx, int l, int r, int val,int pos){    if(arr[pos][idx].right < l||arr[pos][idx].left > r)        return ;    if(arr[pos][idx].left >= l&&arr[pos][idx].right <= r){        arr[pos][idx].sum=(arr[pos][idx].right-arr[pos][idx].left+1)*val;        arr[pos][idx].add=0;        arr[pos][idx].Set=val;        arr[pos][idx].Min=val;        arr[pos][idx].Max=val;        return ;    }    pushDown(idx,pos);    int lc=idx * 2,rc=idx * 2 + 1;    set(lc, l, r, val,pos);    set(rc, l, r, val,pos);    arr[pos][idx].sum = arr[pos][lc].sum + arr[pos][rc].sum;    arr[pos][idx].Min=min(arr[pos][lc].Min,arr[pos][rc].Min);    arr[pos][idx].Max=max(arr[pos][lc].Max,arr[pos][rc].Max);}void add(int idx, int l, int r, int val,int pos){    if(arr[pos][idx].right < l||arr[pos][idx].left > r)        return ;    if(arr[pos][idx].left >= l&&arr[pos][idx].right <= r){        arr[pos][idx].sum += (arr[pos][idx].right - arr[pos][idx].left + 1) * val;        arr[pos][idx].add += val;        arr[pos][idx].Min+=val;        arr[pos][idx].Max+=val;        //printf("pos=%d l=%d r=%d sum=%d add=%d min=%d max=%d\n",pos,arr[pos][idx].left,arr[pos][idx].right,arr[pos][idx].sum,arr[pos][idx].add,          // arr[pos][idx].Min, arr[pos][idx].Max);        return ;    }    pushDown(idx,pos);    int lc=idx * 2,rc=idx * 2 + 1;    add(lc, l, r, val,pos);    add(rc, l, r, val,pos);    arr[pos][idx].sum = arr[pos][lc].sum + arr[pos][rc].sum;    arr[pos][idx].Min=min(arr[pos][lc].Min,arr[pos][rc].Min);    arr[pos][idx].Max=max(arr[pos][lc].Max,arr[pos][rc].Max);    //printf("~~~~~~sum=%d min=%d max=%d\n",arr[pos][idx].sum,arr[pos][idx].Min,arr[pos][idx].Max);}int query(int idx, int l, int r,int pos){    if(arr[pos][idx].right < l||arr[pos][idx].left > r)        return 0;    if(arr[pos][idx].left >= l&&arr[pos][idx].right <= r){        tMin=min(tMin,arr[pos][idx].Min);        tMax=max(tMax,arr[pos][idx].Max);        return arr[pos][idx].sum;    }    pushDown(idx,pos);    //printf("add=%d sum1=%d sum2=%d\n",arr[pos][idx].add,arr[pos][idx * 2].sum , arr[pos][idx * 2 + 1].sum);    int x = query(idx * 2, l, r,pos);    int y = query(idx * 2 + 1, l, r,pos);    //printf("*****%d %d %d %d %d\n",idx,x,y,tMin,tMax);    return x + y;}int main(){    int r,c,m;    while(~scanf("%d %d %d",&r,&c,&m)){        for(int i=1;i<=r;i++){            build(1,1,c,i);        }        while(m--){            int op,x1,x2,y1,y2,v,ans;            scanf("%d",&op);            ans=0;            tMin=INF;            tMax=-INF;            if(op==1){                scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&v);                for(int i=x1;i<=x2;i++){                    add(1,y1,y2,v,i);                }            }else if(op==2){                scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&v);                for(int i=x1;i<=x2;i++){                    set(1,y1,y2,v,i);                }            }else if(op==3){                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);                for(int i=x1;i<=x2;i++){                    int tmp=query(1,y1,y2,i);                    //printf("~~~~%d\n",tmp);                    ans+=tmp;                }                printf("%d %d %d\n",ans,tMin,tMax);            }        }    }return 0;}/**5 5 101 2 2 5 5 32 1 1 3 3 13 2 2 5 53 1 1 3 35 5 101 1 1 3 3 41 2 2 3 3 2**/