BZOJ 1176 CDQ分治

来源：互联网发布：rhinoceros 软件编辑：程序博客网时间：2024/04/29 17:43

光看数据就知道二维数据结构肯定搞不了了，看一眼询问次数就知道离散化也不行，所以就要用一种新的办法——CDQ分治，具体做法如下：将所有询问按照x,y坐标排序，利用CDQ分治的思想，按时间顺序分为左右两半，然后处理左边对右边的影响，用树状数组维护当前点包括它上面所有行的数值的和，将一个询问拆成4个，这道题就可以了

#include<cstdio>#include<cstdlib>#include<cmath>#include<ctime>#include<cstring>#include<string>#include<iostream>#include<iomanip>#include<algorithm>using namespace std;#define QAQ 2147483647int c[2000001];int time_clock[2000001];int n,w;int top;struct caozuo{    int x,y,num,val,ans;    bool operator < (caozuo b) const    {        return x<b.x || (x==b.x && y<b.y);    }}q[200020],nq[200020];bool cmp1(caozuo a,caozuo b){    return a.num<b.num;}bool cmp2(caozuo a,caozuo b){    return a.x<b.x || (a.x==b.x && a.y<b.y);}int tt;inline void add_val(int x,int v){    for(int i=x;i<=n;i+=i&-i)    {        if(time_clock[i]!=tt) c[i]=0;        time_clock[i]=tt;        c[i]+=v;    }}inline int get_ans(int x){    int re=0;    for(int i=x;i;i-=i&-i)        if(time_clock[i]==tt) re+=c[i];    return re;}void add_point(int x,int y,int val){    q[++top].x=x;    q[top].y=y;    q[top].val=val;    q[top].num=top;    if(val==QAQ) q[top].ans+=y*x*w;}void CDQ(int l,int r){    if(l==r) return;    int mid=l+r>>1;    int ll=l;    int rr=mid+1;    for(int i=l;i<=r;i++)    {        if(q[i].num<=mid) nq[ll++]=q[i];        else nq[rr++]=q[i];    }    memcpy(q+l,nq+l,sizeof(q[0])*(r-l+1));    CDQ(l,mid);    CDQ(mid+1,r);    tt++;    int j=l;    for(int i=mid+1;i<=r;i++)    {        while(q[i].x>=q[j].x && j<=mid)        {            if(q[j].val!=QAQ) add_val(q[j].y,q[j].val);            j++;        }        if(q[i].val==QAQ) q[i].ans+=get_ans(q[i].y);    }    int l1=l,l2=mid+1;    for(int i=l;i<=r;i++)    {        if(q[l1]<q[l2]&&l1<=mid||l2>r)            nq[i]=q[l1++];        else            nq[i]=q[l2++];    }    memcpy(q+l,nq+l,sizeof(q[0])*(r-l+1));}int main(){    scanf("%d%d",&w,&n);    int lei;    int x,y,val;    while(1)    {        scanf("%d",&lei);        if(lei==3) break;        if(lei==1)        {            scanf("%d%d%d",&x,&y,&val);            add_point(x,y,val);        }        if(lei==2)        {            int x1,y1,x2,y2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            add_point(x2,y2,QAQ);            add_point(x1-1,y1-1,QAQ);            add_point(x1-1,y2,QAQ);            add_point(x2,y1-1,QAQ);        }    }    sort(q+1,q+1+top,cmp2);    CDQ(1,top);    sort(q+1,q+1+top,cmp1);    for(int i=1;i<=top;i++)    {        if(q[i].val==QAQ)        {            printf("%d\n",q[i].ans+q[i+1].ans-q[i+2].ans-q[i+3].ans);            i+=3;        }    }    return 0;}

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