SPOJ - IITWPC4F 線段樹

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:
3
x 0 1
y 1 2
q 0 0 2 2

Output:
4

題目鏈接

題意：在一個（0<=x、y<=100000）的二維平面上，對點進行操作。初始所有點的值都爲0。我們可以通過一次翻轉將0變成1，1變成0。現在給你n個操作，每個操作有三種可能性。
x l r，將所有x在l到r之間的點翻轉
y l r，將所有y在l到r之間的點翻轉
q x1 y1 x2 y2，詢問在(x1,y1),(x2,y2)這兩個點內的面積內有多少個點值爲1

解題思路：看到區間更新第一個想到的就是線段樹，然而不知道如何向下更新，於是百度了一下，才知道每次update的時候都向下更新一高度，這樣就不用在詢問的時候再遍歷線段樹了。然後最後只要算一下x1到x2，y1到y2之間有多少條值爲1的邊，然後與長度相乘減去重合的就行了。（注意：兩個相交線的點要減兩次，因爲他翻轉兩次會變成0）

#include<cstdio>
#include<cstring>
#define maxn 111111
#define mm 100005
typedef long long ll;
ll n;
char st[5];
struct node{
    ll tree[maxn<<2];
    bool s[maxn<<2];
    void init(){
        memset(tree,0,sizeof(tree));
        memset(s,0,sizeof(0));
    }
    void pushup(ll pos){
        tree[pos]=tree[pos<<1]+tree[pos<<1|1];
    }
    void pushdown(ll pos,ll len)
    {
        if(s[pos])
        {
            tree[pos<<1]=len-len/2-tree[pos<<1];
            tree[pos<<1|1]=len/2-tree[pos<<1|1];
            s[pos<<1]^=1;
            s[pos<<1|1]^=1;
            s[pos]=0;
        }
        return;
    }
    void update(ll L,ll R,ll l,ll r,ll pos){
        if(l>=L&&r<=R){
            tree[pos]=r-l+1-tree[pos];
            s[pos]^=1;
            return;
        }
        pushdown(pos,r-l+1);
        ll mid=(l+r)>>1;
        if(L<=mid)  update(L,R,l,mid,pos<<1);
        if(R>mid)   update(L,R,mid+1,r,pos<<1|1);
        pushup(pos);
    }
    ll query(ll L,ll R,ll l,ll r,ll pos){
        if(L<=l&&r<=R)return tree[pos];
        int mid=(l+r)>>1;
        pushdown(pos,r-l+1);
        int ans=0;
        if(L<=mid) ans+=query(L,R,l,mid,pos<<1);
        if(R>mid) ans+=query(L,R,mid+1,r,pos<<1|1);
        return ans;
    }
}row,col;
ll l,r,x1,x2,y1,y2;
int main(){
    scanf("%lld",&n);
    for(ll i=0;i<n;i++){
        scanf("%s",st);
        if(st[0]=='x'){
            scanf("%lld%lld",&l,&r);
            l++;r++;
            row.update(l,r,1,mm,1);
        }
        else if(st[0]=='y'){
            scanf("%lld%lld",&l,&r);
            l++;r++;
            col.update(l,r,1,mm,1);
        }
        else{
            scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2);
            x1++;x2++;y1++;y2++;
            ll a1=row.query(x1,x2,1,mm,1);
            ll a2=col.query(y1,y2,1,mm,1);
            ll ans=a1*(y2-y1+1)+a2*(x2-x1+1)-2*a1*a2;
            printf("%lld\n",ans);
        }
    }
    return 0;
}