UVA 11992
| 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 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 7Rujia 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!
#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;
}