SPOJ IITWPC4F - Gopu and the Grid Problem （双线段树区间修改区间查询）

Gopu and the Grid Problem

Gopu is interested in the integer co-ordinates of the X-Y plane (0<=x,y<=100000). Each integer coordinate contain a lamp, initially all the lamps are in off mode. Flipping a lamp means switching it on if it is in off mode and vice versa. Maggu will ask gopu 3 type of queries.

Type 1: x l r, meaning: flip all the lamps whose x-coordinate are between l and r (both inclusive) irrespective of the y coordinate.

Type 2: y l r, meaning: flip all the lamps whose y-coordinate are between l and r (both inclusive) irrespective of the x coordinate.

Type 3: q x y X Y, meaning: count the number of lamps which are in ‘on mode’(say this number be A) and ‘off mode’ (say this number be B) in the region where x-coordinate is between x and X(both inclusive) and y-coordinate is between y and Y(both inclusive).

Input

First line contains Q-number of queries that maggu will ask to gopu. (Q <= 10^5)

Then there will be Q lines each containing a query of one of the three type described above.

Output

For each query of 3rd type you need to output a line containing one integer A.

Example

Input:

x 0 1

y 1 2

q 0 0 2 2

 Output:

 4
一开始想的是直接二维线段树或者二维树状数组搞一下，然后一看数据范围，GG。实在没办法看了网上的题解，用两个线段树分别处理行和列，然后查询的时候，再综合一下。
减去重叠的部分。

#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <queue>

#include <vector>

#define inf 0x3f3f3f3f

#define met(a,b) memset(a,b,sizeof a)

#define pb push_back

typedef long long ll;

using namespace std;

const int N = 1e5+;

const int M = ;

typedef long long LL;

int q;

struct Seg {

    int Tree[N << ]; // 标记的行数/列数

    bool lazy[N << ];

    void init() {

        memset(Tree, , sizeof(Tree));

        memset(lazy, , sizeof(lazy));

    }

    inline void pushup(int rt) { Tree[rt] = Tree[rt << ] + Tree[rt <<  | ]; }

    inline void pushdown(int pos,int len)

    {

        if(lazy[pos])

        {

            Tree[pos<<]=len-len/-Tree[pos<<];

            Tree[pos<<|]=len/-Tree[pos<<|];

            lazy[pos<<]^=;

            lazy[pos<<|]^=;

            lazy[pos]=;

        }

        return;

    }

    void update(int L,int R,int l,int r,int pos)

    {

        if(l>=L&&r<=R)

        {

            Tree[pos]=(r-l+)-Tree[pos];

            lazy[pos]^=;

            return;

        }

        int mid=(l+r)>>;

        pushdown(pos,r-l+);

        if(L<=mid) update(L,R,l,mid,pos<<);

        if(mid<R)update(L,R,mid+,r,pos<<|);

        pushup(pos);

    }

    int query(int L,int R,int l,int r,int pos)

    {

        if(L<=l&&r<=R)return Tree[pos];

        int mid=(l+r)>>;

        pushdown(pos,r-l+);

        int ans=;

        if(L<=mid) ans+=query(L,R,l,mid,pos<<);

        if(R>mid) ans+=query(L,R,mid+,r,pos<<|);

        return ans;

    }

}row,col;

int main() {

    char op[];

    int n = N;

    row.init();

    col.init();

    int l,r,val;

    int lx,ly,rx,ry;

    char sign[];

    int m;

    scanf("%d",&m);

    while(m--)

    {

        scanf("%s",sign);

        if(sign[]=='x')

        {

            scanf("%d%d",&l,&r);

            ++l,++r;

            row.update(l,r,,n,);

        }

        else if(sign[]=='y'){

            scanf("%d%d",&l,&r);

            ++l,++r;

            col.update(l,r,,n,);

        }

        else

        {

            scanf("%d%d%d%d",&lx,&ly,&rx,&ry);

            ++lx,++ly,++rx,++ry;

            int nx=row.query(lx,rx,,n,);

            int ny=col.query(ly,ry,,n,);

            ll ans=(ll)(rx-lx+)*ny+(ll)(ry-ly+)*nx-(ll)*nx*ny;

            printf("%lld\n",ans);

        }

    }

}