BZOJ2877 NOI2012魔幻棋盘(二维线段树)

  显然一个序列的gcd=gcd(其差分序列的gcd,序列中第一个数)。于是一维情况直接线段树维护差分序列即可。

  容易想到将该做法拓展到二维。于是考虑维护二维差分,查询时对差分矩阵求矩形的gcd,再对矩形的两个边界求一下原本的gcd即可。

  但这样大概需要三个二维线段树,空间可能不太够。由于查询区域是由一个给定点拓展的,可以改为以该点为中心建差分矩阵,这样剩下部分是一个十字形,可以直接一维线段树维护,就只需要一个二维线段树了。

  注意题面有锅,详见discuss,被坑了一年。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map> 
using namespace std;
#define ll long long
#define N 500010
#define pii pair<int,int>
#define PII pair< pii , pii >
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
ll gcd(ll n,ll m){return m==0?n:gcd(m,n%m);}
ll read()
{
    ll x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,T,X,Y,root[2],cnt[2],CNT,ROOT;
ll BIT[2][N];
vector<ll> a[N];
map<PII,int> id;
struct data{int l,r,L,R;ll gcd;
}tree[2][N<<1],TREE[N*20];
void BIT_add(int op,int n,int x,ll y){while (x<=n) BIT[op][x]+=y,x+=x&-x;}
ll BIT_query(int op,int x){ll s=0;while (x) s+=BIT[op][x],x-=x&-x;return s;}
ll calc(int i,int j)
{
    if (i<X&&j<Y) return a[i][j]-a[i+1][j]-a[i][j+1]+a[i+1][j+1];
    if (i<X&&j>Y) return a[i][j]-a[i+1][j]-a[i][j-1]+a[i+1][j-1];
    if (i>X&&j<Y) return a[i][j]-a[i-1][j]-a[i][j+1]+a[i-1][j+1];
    if (i>X&&j>Y) return a[i][j]-a[i-1][j]-a[i][j-1]+a[i-1][j-1];
    return 0;
}
void add(int &k,int op,int l,int r,int x,ll p)
{
    if (!k) k=++cnt[op];
    if (l==r) {tree[op][k].gcd+=p;return;}
    int mid=l+r>>1;
    if (x<=mid) add(tree[op][k].l,op,l,mid,x,p);
    else add(tree[op][k].r,op,mid+1,r,x,p);
    tree[op][k].gcd=gcd(tree[op][tree[op][k].l].gcd,tree[op][tree[op][k].r].gcd);
}
ll query(int k,int op,int l,int r,int x,int y)
{
    if (x>y||!k) return 0; 
    if (l==x&&r==y) return tree[op][k].gcd;
    int mid=l+r>>1;
    if (y<=mid) return query(tree[op][k].l,op,l,mid,x,y);
    else if (x>mid) return query(tree[op][k].r,op,mid+1,r,x,y);
    else return gcd(query(tree[op][k].l,op,l,mid,x,mid),query(tree[op][k].r,op,mid+1,r,mid+1,y));
}
void BUILD(int &k,int l,int r,int x)
{
    id[make_pair(make_pair(x,x),make_pair(l,r))]=k=++CNT;
    if (l==r) {TREE[k].gcd=calc(x,l);return;}
    int mid=l+r>>1;
    BUILD(TREE[k].l,l,mid,x);
    BUILD(TREE[k].r,mid+1,r,x);
    TREE[k].gcd=gcd(TREE[TREE[k].l].gcd,TREE[TREE[k].r].gcd);
}
void BUILD2(int &k,int l,int r,int u,int x,int y)
{
    if (!k) id[make_pair(make_pair(l,r),make_pair(x,y))]=k=++CNT;
    TREE[k].L=id[make_pair(make_pair(l,u),make_pair(x,y))];
    TREE[k].R=id[make_pair(make_pair(u+1,r),make_pair(x,y))];
    TREE[k].gcd=gcd(TREE[TREE[k].L].gcd,TREE[TREE[k].R].gcd);
    if (x==y) return;
    int mid=x+y>>1;
    BUILD2(TREE[k].l,l,r,u,x,mid);
    BUILD2(TREE[k].r,l,r,u,mid+1,y);
}
void update(int k,int l,int r,int x)
{
    TREE[k].gcd=gcd(TREE[TREE[k].L].gcd,TREE[TREE[k].R].gcd);
    if (l==r) return;
    int mid=l+r>>1;
    if (x<=mid) update(TREE[k].l,l,mid,x);
    else update(TREE[k].r,mid+1,r,x);
}
void build(int &k,int l,int r)
{
    if (l==r) {BUILD(k,0,m+1,l);return;}
    id[make_pair(make_pair(l,r),make_pair(0,m+1))]=k=++CNT;
    int mid=l+r>>1;
    build(TREE[k].L,l,mid);
    build(TREE[k].R,mid+1,r);
    BUILD2(k,l,r,(l+r>>1),0,m+1);
}
void ADD(int &k,int l,int r,int x,ll p)
{
    if (l==r) {TREE[k].gcd+=p;return;}
    int mid=l+r>>1;
    if (x<=mid) ADD(TREE[k].l,l,mid,x,p);
    else ADD(TREE[k].r,mid+1,r,x,p);
    TREE[k].gcd=gcd(TREE[TREE[k].l].gcd,TREE[TREE[k].r].gcd);
}
void Add(int k,int l,int r,int x,int y,ll p)
{
    if (l==r) {ADD(k,0,m+1,y,p);return;}
    int mid=l+r>>1;
    if (x<=mid) Add(TREE[k].L,l,mid,x,y,p);
    else Add(TREE[k].R,mid+1,r,x,y,p);
    update(k,0,m+1,y);
}
ll QUERY(int k,int l,int r,int x,int y)
{
    if (l==x&&r==y) return TREE[k].gcd;
    int mid=l+r>>1;
    if (y<=mid) return QUERY(TREE[k].l,l,mid,x,y);
    else if (x>mid) return QUERY(TREE[k].r,mid+1,r,x,y);
    else return gcd(QUERY(TREE[k].l,l,mid,x,mid),QUERY(TREE[k].r,mid+1,r,mid+1,y));
}
ll Query(int k,int l,int r,int xl,int xr,int yl,int yr)
{
    if (xl>xr||yl>yr) return 0;
    if (l==xl&&r==xr) return QUERY(k,0,m+1,yl,yr);
    int mid=l+r>>1;
    if (xr<=mid) return Query(TREE[k].L,l,mid,xl,xr,yl,yr);
    else if (xl>mid) return Query(TREE[k].R,mid+1,r,xl,xr,yl,yr);
    else return gcd(Query(TREE[k].L,l,mid,xl,mid,yl,yr),Query(TREE[k].R,mid+1,r,mid+1,xr,yl,yr));
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj2877.in","r",stdin);
    freopen("bzoj2877.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read(),X=read(),Y=read(),T=read();
    for (int j=0;j<=m+1;j++) a[0].push_back(0);
    for (int i=1;i<=n;i++)
    {
        a[i].push_back(0);
        for (int j=1;j<=m;j++)
        a[i].push_back(read());
        a[i].push_back(0);
    }
    for (int j=0;j<=m+1;j++) a[n+1].push_back(0);
    build(ROOT,0,n+1);
    for (int i=1;i<=n;i++)
    BIT_add(0,n,i,a[i][Y]-a[i-1][Y]),add(root[0],0,1,n,i,a[i][Y]-a[i-1][Y]);
    for (int j=1;j<=m;j++)
    BIT_add(1,m,j,a[X][j]-a[X][j-1]),add(root[1],1,1,m,j,a[X][j]-a[X][j-1]);
    while (T--)
    {
        int op=read();
        if (op==0)
        {
            int up=read(),left=read(),down=read(),right=read();
            ll ans=gcd(BIT_query(0,X-up),query(root[0],0,1,n,X-up+1,X+down));
            ans=gcd(ans,gcd(BIT_query(1,Y-left),query(root[1],1,1,m,Y-left+1,Y+right)));
            ans=gcd(ans,Query(1,0,n+1,X-up,X-1,Y-left,Y-1));
            ans=gcd(ans,Query(1,0,n+1,X-up,X-1,Y+1,Y+right));
            ans=gcd(ans,Query(1,0,n+1,X+1,X+down,Y-left,Y-1));
            ans=gcd(ans,Query(1,0,n+1,X+1,X+down,Y+1,Y+right));
            printf(LL,abs(ans));
        }
        else
        {
            int xl=read(),yl=read(),xr=read(),yr=read();ll c=read();
            if (yl<=Y&&Y<=yr)
            {
                BIT_add(0,n,xl,c),add(root[0],0,1,n,xl,c);
                if (xr<n) BIT_add(0,n,xr+1,-c),add(root[0],0,1,n,xr+1,-c);
            }
            if (xl<=X&&X<=xr)
            {
                BIT_add(1,m,yl,c),add(root[1],1,1,m,yl,c);
                if (yr<m) BIT_add(1,m,yr+1,-c),add(root[1],1,1,m,yr+1,-c);
            }
            if (xl<=X&&yl<=Y)
            {
                if (xr<X&&yr<Y) Add(1,0,n+1,xr,yr,c);
                if (xr<X) Add(1,0,n+1,xr,yl-1,-c);
                if (yr<Y) Add(1,0,n+1,xl-1,yr,-c);
                Add(1,0,n+1,xl-1,yl-1,c);
            }
            if (xr>=X&&yl<=Y)
            {
                if (xl>X&&yr<Y) Add(1,0,n+1,xl,yr,c);
                if (xl>X) Add(1,0,n+1,xl,yl-1,-c);
                if (yr<Y) Add(1,0,n+1,xr+1,yr,-c);
                Add(1,0,n+1,xr+1,yl-1,c);
            }
            if (xl<=X&&yr>=Y)
            {
                if (xr<X&&yl>Y) Add(1,0,n+1,xr,yl,c);
                if (xr<X) Add(1,0,n+1,xr,yr+1,-c);
                if (yl>Y) Add(1,0,n+1,xl-1,yl,-c);
                Add(1,0,n+1,xl-1,yr+1,c);
            }
            if (xr>=X&&yr>=Y)
            {
                if (xl>X&&yl>Y) Add(1,0,n+1,xl,yl,c);
                if (xl>X) Add(1,0,n+1,xl,yr+1,-c);
                if (yl>Y) Add(1,0,n+1,xr+1,yl,-c);
                Add(1,0,n+1,xr+1,yr+1,c);
            }
        }
    }
    return 0;
}

 

  

 

posted @ 2019-02-01 17:51  Gloid  阅读(219)  评论(0编辑  收藏  举报