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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.
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.
For each type-3 query, print the summation, min and max.
线段树，"区间加"和"区间修改"操作
由于矩阵最多有20行,所以可以将每行合到一个序列上
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEM2(a,i) memset(a,i,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXR (20+1)
#define MAXN (8000000+10)
#define MAXQ (20000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
class SegmentTree
{
ll a[MAXN],minv[MAXN],sumv[MAXN],maxv[MAXN],addv[MAXN],setv[MAXN];
int n;
public:
SegmentTree(){ }
SegmentTree(int _n):n(_n){ }
void mem(int _n)
{
n=_n;
For(i,4*n+3) a[i]=minv[i]=sumv[i]=maxv[i]=addv[i]=0,setv[i]=-1;
}
void maintain(int o,int L,int R)
{
sumv[o]=maxv[o]=minv[o]=0;
if (L<R) //只考虑左右子树
{
sumv[o]=sumv[Lson]+sumv[Rson];
minv[o]=min(minv[Lson],minv[Rson]);
maxv[o]=max(maxv[Lson],maxv[Rson]);
} //只考虑add操作
if (setv[o]>0) sumv[o]=setv[o]*(R-L+1),minv[o]=maxv[o]=setv[o];
minv[o]+=addv[o];maxv[o]+=addv[o];sumv[o]+=addv[o]*(R-L+1);
}
int y1,y2,v;
void update(int o,int L,int R) //y1,y2,v
{
if (y1<=L&&R<=y2) {
addv[o]+=v;
}
else{
pushdown(o);
int M=(R+L)>>1;
if (y1<=M) update(Lson,L,M); else maintain(Lson,L,M);
if (M< y2) update(Rson,M+1,R); else maintain(Rson,M+1,R);
}
maintain(o,L,R);
}
void update2(int o,int L,int R)
{
if (y1<=L&&R<=y2) {
setv[o]=v;addv[o]=0;
}
else{
pushdown(o);
int M=(R+L)>>1;
if (y1<=M) update2(Lson,L,M); else maintain(Lson,L,M); //维护pushodown，再次maintain
if (M< y2) update2(Rson,M+1,R); else maintain(Rson,M+1,R);
}
maintain(o,L,R);
}
void pushdown(int o)
{
if (setv[o]>=0)
{
setv[Lson]=setv[Rson]=setv[o];
addv[Lson]=addv[Rson]=0;
setv[o]=-1;
}
if (addv[o])
{
addv[Lson]+=addv[o];
addv[Rson]+=addv[o];
addv[o]=0;
}
}
void query2(int o,int L,int R,ll add)
{
if (setv[o]>=0)
{
_sum+=(setv[o]+addv[o]+add)*(min(R,y2)-max(L,y1)+1);
_min=min(_min,setv[o]+addv[o]+add);
_max=max(_max,setv[o]+addv[o]+add);
} else if (y1<=L&&R<=y2)
{
_sum+=sumv[o]+add*(R-L+1);
_min=min(_min,minv[o]+add);
_max=max(_max,maxv[o]+add);
} else {
int M=(L+R)>>1;
if (y1<=M) query2(Lson,L,M,add+addv[o]);
if (M< y2) query2(Rson,M+1,R,add+addv[o]);
}
}
ll _min,_max,_sum;
void add(ll v,int l,int r)
{
y1=l,y2=r;this->v=v;
update(1,1,n);
}
void set(ll v,int l,int r)
{
y1=l,y2=r;this->v=v;
update2(1,1,n);
}
ll ask(int l,int r,int b=0)
{
_sum=0,_min=INF,_max=-1;
y1=l,y2=r;
query2(1,1,n,0);
switch(b)
{
case 1:return _sum;
case 2:return _min;
case 3:return _max;
default:break;
}
}
//先set后add
}S;
int main()
{
// freopen("uva11992.in","r",stdin);
// freopen(".out","w",stdout);
int R,n,Q;
while (~scanf("%d%d%d",&R,&n,&Q))
{
S.mem(R*n);
int p,x1,y1,x2,y2;
while(Q--)
{
scanf("%d%d%d%d%d",&p,&x1,&y1,&x2,&y2);
ll v;
if (p<=2) scanf("%lld",&v);
if (p==1) {
Fork(i,x1,x2) S.add(v,(i-1)*n+y1,(i-1)*n+y2);
}
else if (p==2) {
Fork(i,x1,x2) S.set(v,(i-1)*n+y1,(i-1)*n+y2);
} else {
ll s=0,mi=INF,ma=-1;
Fork(i,x1,x2)
{
S.ask((i-1)*n+y1,(i-1)*n+y2);
s+=S._sum;mi=min(mi,S._min);ma=max(ma,S._max);
}
printf("%lld %lld %lld\n",s,mi,ma);
}
}
}
return 0;
}
UVA 11992(Fast Matrix Operations-线段树区间加&改)[Template:SegmentTree]
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原文地址：http://blog.csdn.net/nike0good/article/details/45695145
