BZOJ1935 园丁的烦恼

一个二维偏序的问题，学过了三维偏序cdq分治之后觉得这个题非常的水。只需按一维排序之后再用树状数组操作即可。——by VANE

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5;
struct qry
{
    int x,y,id;
}q[N<<2];
struct tree
{
    int x,y;
}t[N];
int s[N*5],n,m,tmp[N*5],num,vis_tim,ans[N*5];
bool cmp(qry a,qry b)
{
    return a.x<b.x;
}
bool cmpp(tree a,tree b)
{
    return a.x<b.x;
}
void add(int x)
{
    for(;x<=num;x+=x&-x)
        s[x]++;
}
int query(int x)
{
    int res=0;
    for(;x;x-=x&-x)
        res+=s[x];
    return res;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
        scanf("%d%d",&t[i].x,&t[i].y);
    for(int i=1;i<=m;++i)
    {
        int x,y,xx,yy;scanf("%d%d%d%d",&x,&y,&xx,&yy);
        int pos=(i-1)*4;
        q[pos+1].x=xx;q[pos+1].y=yy;q[pos+1].id=pos+1;
        q[pos+2].x=x-1;q[pos+2].y=y-1;q[pos+2].id=pos+2;
        q[pos+3].x=xx;q[pos+3].y=y-1;q[pos+3].id=pos+3;
        q[pos+4].x=x-1;q[pos+4].y=yy;q[pos+4].id=pos+4;
    }
    for(int i=1;i<=n;++i)
    tmp[i]=t[i].y;
    for(int i=1;i<=m*4;++i)
    tmp[i+n]=q[i].y;
    sort(tmp+1,tmp+1+n+m*4);
    num=unique(tmp+1,tmp+1+n+4*m)-tmp-1;
    for(int i=1;i<=n;++i) t[i].y=lower_bound(tmp+1,tmp+1+num,t[i].y)-tmp;
    for(int i=1;i<=m*4;++i) q[i].y=lower_bound(tmp+1,tmp+1+num,q[i].y)-tmp;
    sort(q+1,q+1+4*m,cmp);
    sort(t+1,t+1+n,cmpp);
    int i=1,j=1;
    while(j<=4*m)
    {
        int h=q[j].x;
        while(t[i].x<=h&&i<=n) add(t[i].y),++i;
        while(j<=4*m&&q[j].x==h)
        {
            ans[q[j].id]+=query(q[j].y);
            ++j;
        }
    }
    for(int i=1;i<=4*m;i+=4)
        printf("%d\n",ans[i]+ans[i+1]-ans[i+2]-ans[i+3]);
}

 二维偏序用cdq我是不是有病，我就是要写cdq 15000ms卡过去了

By：大奕哥

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 struct node
 4 {
 5     int x,y,w,id,f;
 6     bool operator <(const node &b)const
 7     {
 8         return x==b.x?y<b.y:x<b.x;
 9     }
10 }q[2500005];
11 int cnt,maxr,tim;
12 int t[10000005],v[10000005],ans[2500005],pos[500005];
13 inline int lowbit(int x){return x&(-x);}
14 void add(int x,int w)
15 {
16     for(;x<=maxr;x+=lowbit(x))
17     {
18         if(v[x]!=tim)
19         {
20             v[x]=tim;t[x]=w;
21         }
22         else t[x]+=w;
23     }
24 }
25 int query(int x)
26 {
27     int an=0;
28     for(;x;x-=lowbit(x))
29     if(v[x]==tim)
30     an+=t[x];
31     return an;
32 }
33 void cdq(int l,int r)
34 {
35     if(l==r)return;
36     int mid=(l+r)>>1;
37     cdq(l,mid);cdq(mid+1,r);
38     sort(q+l,q+mid+1);sort(q+mid+1,q+r+1);
39     tim++;
40     int i=l,j=mid+1;
41     while(j<=r)
42     {
43         while(q[i].f==2&&i<=mid)++i;
44         while(q[j].f==1&&j<=r)++j;
45         if(q[i].x<=q[j].x&&i<=mid)add(q[i].y,q[i].w),++i;
46         else if(j<=r)ans[q[j].id]+=query(q[j].y),++j;
47     }
48 }
49 int main()
50 {
51     int n,m;
52     scanf("%d%d",&n,&m);
53     for(int i=1;i<=n;++i)
54     {
55         int x,y;
56         scanf("%d%d",&x,&y);x++;y++;
57         q[++cnt].f=1;q[cnt].x=x;q[cnt].y=y;q[cnt].w=1;q[cnt].id=cnt;maxr=max(maxr,y);
58     }
59     for(int i=1;i<=m;++i)
60     {
61         int x1,x2,y1,y2;
62         scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++;x2++;y1++;y2++;
63         pos[i]=cnt;maxr=max(maxr,max(y1,y2));
64         q[++cnt].f=2;q[cnt].x=x2;q[cnt].y=y2;q[cnt].id=cnt;
65         q[++cnt].f=2;q[cnt].x=x1-1;q[cnt].y=y1-1;q[cnt].id=cnt;
66         q[++cnt].f=2;q[cnt].x=x1-1;q[cnt].y=y2;q[cnt].id=cnt;
67         q[++cnt].f=2;q[cnt].x=x2;q[cnt].y=y1-1;q[cnt].id=cnt;        
68     }
69     cdq(1,cnt);
70     for(int i=1;i<=m;++i)
71     printf("%d\n",ans[pos[i]+1]+ans[pos[i]+2]-ans[pos[i]+3]-ans[pos[i]+4]);
72     return 0;
73 }

 

posted @ 2017-12-18 21:24  大奕哥&VANE  阅读(161)  评论(0)  编辑  收藏  举报