[BZOJ2877][NOI2012]魔幻棋盘(二维线段树)

https://blog.sengxian.com/solutions/bzoj-2877

注意二维线段树的upd()也是一个O(log n)的函数(pushdown()应该也是但没写过)。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #define Ls (x<<1)
 4 #define Rs (Ls|1)
 5 #define Lson Ls,L,mid
 6 #define Rson Rs,mid+1,R
 7 #define lson ls[x],L,mid
 8 #define rson rs[x],mid+1,R
 9 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
10 typedef long long ll;
11 using namespace std;
12 
13 const int N=500010,M=10000010;
14 ll c,sm[M],a[N];
15 int n,m,x,y,T,nd,op,x1,x2,y1,y2,rt[N<<2],ls[M],rs[M];
16 
17 int F(int x,int y){ return (x-1)*m+y; }
18 ll gcd(ll a,ll b){ return b ? gcd(b,a%b) : a; }
19 
20 void mdfy(int &x,int L,int R,int pos,ll k){
21     if (!x) x=++nd;
22     if (L==R){ sm[x]+=k; return; }
23     int mid=(L+R)>>1;
24     if (pos<=mid) mdfy(lson,pos,k); else mdfy(rson,pos,k);
25     sm[x]=gcd(sm[ls[x]],sm[rs[x]]);
26 }
27 
28 void upd(int &x,int L,int R,int pos,int lp,int rp){
29     if (!x) x=++nd;
30     if (L==R){ sm[x]=gcd(sm[lp],sm[rp]); return; }
31     int mid=(L+R)>>1;
32     if (pos<=mid) upd(lson,pos,ls[lp],ls[rp]); else upd(rson,pos,rs[lp],rs[rp]);
33     sm[x]=gcd(sm[ls[x]],sm[rs[x]]);
34 }
35 
36 void mdfx(int x,int L,int R,int x1,int y1,ll k){
37     if (L==R){ mdfy(rt[x],0,m,y1,k); return; }
38     int mid=(L+R)>>1;
39     if (x1<=mid) mdfx(Lson,x1,y1,k); else mdfx(Rson,x1,y1,k);
40     upd(rt[x],0,m,y1,rt[Ls],rt[Rs]);
41 }
42 
43 ll quey(int x,int L,int R,int l,int r){
44     if (L==l && r==R) return sm[x];
45     int mid=(L+R)>>1;
46     if (r<=mid) return quey(lson,l,r);
47     else if (l>mid) return quey(rson,l,r);
48         else return gcd(quey(lson,l,mid),quey(rson,mid+1,r));
49 }
50 
51 ll quex(int x,int L,int R,int l,int r,int y1,int y2){
52     if (L==l && r==R) return quey(rt[x],0,m,y1,y2);
53     int mid=(L+R)>>1;
54     if (r<=mid) return quex(Lson,l,r,y1,y2);
55     else if (l>mid) return quex(Rson,l,r,y1,y2);
56         else return gcd(quex(Lson,l,mid,y1,y2),quex(Rson,mid+1,r,y1,y2));
57 }
58 
59 int main(){
60     freopen("chess.in","r",stdin);
61     freopen("chess.out","w",stdout);
62     scanf("%d%d%d%d%d",&n,&m,&x,&y,&T); int r1=(n+1)*4+1,r2=r1+1;
63     rep(i,1,n) rep(j,1,m) scanf("%lld",&a[F(i,j)]);
64     rep(i,1,n-1) rep(j,1,m-1) mdfx(1,0,n,i,j,a[F(i+1,j+1)]-a[F(i+1,j)]-a[F(i,j+1)]+a[F(i,j)]);
65     rep(i,1,n-1) mdfy(rt[r1],0,n,i,a[F(i+1,y)]-a[F(i,y)]);
66     rep(i,1,m-1) mdfy(rt[r2],0,m,i,a[F(x,i+1)]-a[F(x,i)]);
67     while (T--){
68         scanf("%d%d%d%d%d",&op,&x1,&y1,&x2,&y2);
69         if (op==0){
70             ll t1=(x1+x2) ? quey(rt[r1],0,n,x-x1,x+x2-1) : 0;
71             ll t2=(y1+y2) ? quey(rt[r2],0,m,y-y1,y+y2-1) : 0;
72             ll t3=((x1+x2)&&(y1+y2)) ? quex(1,0,n,x-x1,x+x2-1,y-y1,y+y2-1) : 0;
73             printf("%lld\n",abs(gcd(gcd(t1,t2),gcd(t3,a[F(x,y)]))));
74         }else{
75             scanf("%lld",&c);
76             mdfx(1,0,n,x1-1,y1-1,c); mdfx(1,0,n,x1-1,y2,-c);
77             mdfx(1,0,n,x2,y1-1,-c); mdfx(1,0,n,x2,y2,c);
78             if (y1<=y && y2>=y) mdfy(rt[r1],0,n,x1-1,c),mdfy(rt[r1],0,n,x2,-c);
79             if (x1<=x && x2>=x) mdfy(rt[r2],0,m,y1-1,c),mdfy(rt[r2],0,m,y2,-c);
80             if (x1<=x && x2>=x && y1<=y && y2>=y) a[F(x,y)]+=c;
81         }
82     }
83     return 0;
84 }

 

posted @ 2019-03-20 09:23  HocRiser  阅读(285)  评论(0编辑  收藏  举报