UVA 11992

Fast Matrix Operations

Time Limit:5000MS Memory Limit:Unknown 64bit IO Format: %lld & %llu

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Description

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!

蛋碎的一题。。线段树的区间合并类的。   要用到多维线段树，用一维数组表示，每4*c（c表示矩阵列数）表示一个线段树。

 对于最小，最大，还有区间和，直接简单的合并即可。 将区间所有的和加V，与设置V  分别开两个LAZY标记  add,和cov。当给区间加V时，直接加在区间的add上。  当给区间 设置为V时，直接 把cov设置成v，同时改变add为0.

在处理的时候因为运算优先级。。一直查不出哪里有错。。

T[t+id<<1|1]    //不能这样写。。’+‘运算优先级比 '<<'高。 要写成  T[t+(id<<1|1)].....

过了样例后，直接交了。居然1A。。

#include<cstdio>#include<vector>#include<algorithm>#include<cstring>using namespace std;int r,c,m;const int maxn = 1000010;const int inf = 1000000009;int amax[22],amin[22],asum[22];struct node{    int l,r,sum,min,max,add,cov;}T[maxn*5];void build(int i,int id,int l,int r){    int t=4*i*c;    T[id+t].l=l;  T[id+t].r=r;  T[id+t].add=0;  T[id+t].cov=-1;T[id+t].sum=T[id+t].min=T[id+t].max=0;    if(l==r)   return ;    int m=(l+r)>>1;    build(i,id<<1,l,m);build(i,id<<1|1,m+1,r);}void pushDown(int k,int id){    int t=4*k*c;    if(T[t+id].cov!=-1){        if(T[t+id].l!=T[t+id].r){        T[t+(id<<1)].add=T[t+(id<<1|1)].add=0;        T[t+(id<<1)].cov=T[t+(id<<1|1)].cov=T[t+id].cov;        T[t+(id<<1)].max=T[t+(id<<1|1)].max=T[t+id].cov;        T[t+(id<<1)].min=T[t+(id<<1|1)].min=T[t+id].cov;        T[t+(id<<1)].sum=(T[t+(id<<1)].r-T[t+(id<<1)].l+1)*T[t+id].cov;        T[t+(id<<1|1)].sum=(T[t+(id<<1|1)].r-T[t+(id<<1|1)].l+1)*T[t+id].cov;        }        T[t+id].cov=-1;    }    int tt=T[t+id].add;    if(T[t+id].add>0){        if(T[t+id].l!=T[t+id].r){        T[t+(id<<1)].add+=tt; T[t+(id<<1|1)].add+=tt;        T[t+(id<<1)].max+=tt; T[t+(id<<1|1)].max+=tt;        T[t+(id<<1)].min+=tt; T[t+(id<<1|1)].min+=tt;        T[t+(id<<1)].sum+=tt*(T[t+(id<<1)].r-T[t+(id<<1)].l+1);        T[t+(id<<1|1)].sum+=tt*(T[t+(id<<1|1)].r-T[t+(id<<1|1)].l+1);        }        T[t+id].add=0;    }}void pushUp(int k,int id){    int t=4*k*c;    T[t+id].max=max(T[t+(id<<1)].max,T[t+(id<<1|1)].max);    T[t+id].min=min(T[t+(id<<1)].min,T[t+(id<<1|1)].min);    T[t+id].sum=T[t+(id<<1)].sum+T[t+(id<<1|1)].sum;}void update_add(int k,int id,int l,int r,int val){     int t=4*k*c+id;     if(T[t].l==l&&T[t].r==r){        T[t].add+=val;        T[t].max+=val;        T[t].min+=val;        T[t].sum+=val*(r-l+1);        return ;     }     pushDown(k,id);     int m=(T[t].l+T[t].r)>>1;     if(m>=r)   update_add(k,id<<1,l,r,val);     else if(m<l) update_add(k,id<<1|1,l,r,val);     else{          update_add(k,id<<1,l,m,val);          update_add(k,id<<1|1,m+1,r,val);     }     pushUp(k,id);}void update_set(int k,int id,int l,int r,int val){     int t=4*k*c+id;     if(T[t].l==l&&T[t].r==r){        T[t].cov=val;  T[t].add=0;        T[t].max=val;        T[t].min=val;        T[t].sum=val*(r-l+1);        return ;     }     pushDown(k,id);     int m=(T[t].l+T[t].r)>>1;     if(m>=r)   update_set(k,id<<1,l,r,val);     else if(m<l) update_set(k,id<<1|1,l,r,val);     else{          update_set(k,id<<1,l,m,val);          update_set(k,id<<1|1,m+1,r,val);     }     pushUp(k,id);}void query(int k,int id,int l,int r){    int t=4*k*c+id;    if(T[t].l==l&&T[t].r==r){         asum[k]+=T[t].sum;         amax[k]=max(amax[k],T[t].max);         amin[k]=min(amin[k],T[t].min);         return ;    }     pushDown(k,id);     int m=(T[t].l+T[t].r)>>1;     if(m>=r)   query(k,id<<1,l,r);     else if(m<l) query(k,id<<1|1,l,r);     else{          query(k,id<<1,l,m);          query(k,id<<1|1,m+1,r);     }     pushUp(k,id);}int main(){    while(scanf("%d%d%d",&r,&c,&m)!=EOF){         for(int i=1;i<=r;i++)         build(i,1,1,c+1);         int id,x1,y1,x2,y2,v;         while(m--){             scanf("%d",&id);             if(id==1){                 scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);                 for(int i=x1;i<=x2;i++)                    update_add(i,1,y1,y2,v);             }             else if(id==2){                 scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);                 for(int i=x1;i<=x2;i++)                    update_set(i,1,y1,y2,v);             }             else {                 memset(asum,0,sizeof(asum));                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                 int sum=0,rmax=-inf,rmin=inf;                 for(int i=x1;i<=x2;i++) amax[i]=-inf,amin[i]=inf;                 for(int i=x1;i<=x2;i++){                     query(i,1,y1,y2);                     sum+=asum[i];                     rmax=max(rmax,amax[i]);                     rmin=min(rmin,amin[i]);                 }                 printf("%d %d %d\n",sum,rmin,rmax);             }         }    }    return 0;}