【UVA11992】Fast Matrix Operations——二维线段树

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAX = 41000;

struct node {

int sum,maxn,minn;
int op,lazy;
int lson,rson;
void Clear() {
sum = 0; maxn = 0; minn = INF;
op = lazy = 0;
}
node operator + (const node &a)const {
node ans;
ans.Clear();
ans.sum = sum+a.sum;
ans.minn = min(minn,a.minn);
ans.maxn = max(maxn,a.maxn);
return ans;
}
}tr[42][MAX*5];

void pushup(int index,int st) {
tr[index][st].sum = tr[index][st<<1].sum+tr[index][st<<1|1].sum;
tr[index][st].minn = min(tr[index][st<<1].minn,tr[index][st<<1|1].minn);
tr[index][st].maxn = max(tr[index][st<<1].maxn,tr[index][st<<1|1].maxn);
}

void pushdown1(node &a ,node &b) {
b.sum +=(b.rson - b.lson+1)*a.lazy;
b.maxn += a.lazy;
b.minn += a.lazy;
b.op = b.op == 2?2:1;
b.lazy += a.lazy;
}
void pushdown2(node &a,node &b) {
b.sum = (b.rson - b.lson +1)*a.lazy;
b.maxn = b.minn = a.lazy;
b.op = 2;
b.lazy = a.lazy;
}
void pushdown(int index,int st) {//比较恶心的地方，向下lazy更新
if(tr[index][st].lson == tr[index][st].rson || tr[index][st].op == 0) return ;
if(tr[index][st].op == 1) {
pushdown1(tr[index][st],tr[index][st<<1]);
pushdown1(tr[index][st],tr[index][st<<1|1]);
}
else {
pushdown2(tr[index][st],tr[index][st<<1]);
pushdown2(tr[index][st],tr[index][st<<1|1]);
}
tr[index][st].lazy = tr[index][st].op = 0;
}

void Build(int index,int L,int R,int st) {
tr[index][st].lson = L;
tr[index][st].rson = R;
tr[index][st].op = tr[index][st].lazy = 0;
if(L == R) {
tr[index][st].maxn = tr[index][st].sum = 0;
tr[index][st].minn = 0;
return ;
}
int mid = (L+R)>>1;
Build(index,L,mid,st<<1);
Build(index,mid+1,R,st<<1|1);
pushup(index,st);
}

void Update(int index,int st,int l,int r,int val,int op) {//区间更新，找清两种操作之间的关系
pushdown(index,st);
if(l == tr[index][st].lson && r == tr[index][st].rson) {
if(op == 1) {
tr[index][st].sum += (r-l+1)*val;
tr[index][st].minn += val;
tr[index][st].maxn += val;
tr[index][st].op = tr[index][st].op==2?2:1;
tr[index][st].lazy += val;
}
else {
tr[index][st].sum =  (r-l+1)*val;
tr[index][st].maxn = tr[index][st].minn = val;
tr[index][st].op = op;
tr[index][st].lazy = val;
}
return ;
}
int mid = (tr[index][st].lson+tr[index][st].rson)>>1;
if(r<=mid) Update(index,st<<1,l,r,val,op);
else if(l>mid) Update(index,st<<1|1,l,r,val,op);
else {
Update(index,st<<1,l,mid,val,op);
Update(index,st<<1|1,mid+1,r,val,op);
}
pushup(index,st);
}

node Query(int index,int st,int l,int r) {//查询操作
pushdown(index,st);
node ans; ans.Clear();
if(l == tr[index][st].lson && tr[index][st].rson == r) return tr[index][st];
int mid = (tr[index][st].lson + tr[index][st].rson ) >>1;
if(r<=mid) ans = ans + Query(index,st<<1,l,r);
else if(l>mid)  ans = ans + Query(index,st<<1|1,l,r);
else {
ans = ans + Query(index,st<<1,l,mid);
ans = ans + Query(index,st<<1|1,mid+1,r);
}
pushup(index,st);//要及时的更新上去
return ans;
}

int main() {
int n,m,q,op,x1,y1,x2,y2,v;
while(~scanf("%d %d %d",&n,&m,&q)) {
for(int i = 1; i <= n; i++) Build(i,1,m,1);
while(q--) {
scanf("%d %d %d %d %d",&op,&x1,&y1,&x2,&y2);
if(op!=3) {
scanf("%d",&v);
for(int i = x1;i<=x2;i++) Update(i,1,y1,y2,v,op);
}
else {
node ans; ans.Clear();
for(int i = x1;i<=x2;i++) ans =ans +  Query(i,1,y1,y2);
printf("%d %d %d\n",ans.sum,ans.minn,ans.maxn);
}
}
}
return 0;
}
/*
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:
45 0 5
78 5 7
69 2 7
39 2 7
*/