很明显直接二维是作死的。。所以给按X点排下序,当然离线处理询问时差分的点会方便些。。

对Y坐标就用BIT维护就行。。

但由于坐标比较大,所以还要离散化= =

处理起来略麻烦但代码并不难写

 1 //#include<bits/stdc++.h>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<iostream>
 7 #define inc(i,l,r) for(int i=l;i<=r;i++)
 8 #define dec(i,l,r) for(int i=l;i>=r;i--)
 9 #define link(x) for(edge *j=h[x];j;j=j->next)
10 #define mem(a) memset(a,0,sizeof(a))
11 #define ll long long
12 #define succ(x) (1<<x)
13 #define lowbit(x) (x&(-x))
14 #define NM 500000+5
15 using namespace std;
16 int read(){
17     int x=0,f=1;char ch=getchar();
18     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
19     while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
20     return x*f;
21 }
22 struct tmp{
23     int x,y,id,f;
24 }a[5*NM];
25 bool cmp(tmp x,tmp y){
26     return x.x<y.x||(x.x==y.x&&x.id<y.id);
27 }
28 int _x,_y,__x,__y,b[5*NM],c[NM],ans[NM][5],tot,n,m,_t;
29 void upd(int x,int t){
30     for(;x<=n;x+=lowbit(x))c[x]+=t;
31 }
32 int sum(int x){
33     int s=0;
34     for(;x;x-=lowbit(x))s+=c[x];
35     return s;
36 }
37 int main(){
38     freopen("data.in","r",stdin);
39     n=read();m=read();
40     inc(i,1,n){
41         a[++tot].x=read();a[tot].y=read();b[tot]=a[i].y;
42     }
43     inc(i,1,m){
44         _x=read()-1;_y=read()-1;__x=read();__y=read();
45         a[++tot].x=_x;a[tot].y=_y;b[tot]=_y;a[tot].id=i;a[tot].f=1;
46         a[++tot].x=_x;a[tot].y=__y;b[tot]=__y;a[tot].id=i;a[tot].f=2;
47         a[++tot].x=__x;a[tot].y=_y;b[tot]=_y;a[tot].id=i;a[tot].f=3;
48         a[++tot].x=__x;a[tot].y=__y;b[tot]=__y;a[tot].id=i;a[tot].f=4;
49     }
50     sort(b+1,b+tot+1);
51     _t=unique(b+1,b+tot+1)-b-1;
52     inc(i,1,tot)a[i].y=lower_bound(b+1,b+_t+1,a[i].y)-b;
53     sort(a+1,a+tot+1,cmp);
54 /*    inc(i,1,tot)printf("%d ",a[i].x);printf("\n");
55     inc(i,1,tot)printf("%d ",a[i].y);printf("\n");
56     inc(i,1,tot)printf("%d ",a[i].id);printf("\n");*/
57     inc(i,1,tot)
58     if(a[i].id)ans[a[i].id][a[i].f]=sum(a[i].y);
59     else upd(a[i].y,1);
60     inc(i,1,m)printf("%d\n",ans[i][1]+ans[i][4]-ans[i][2]-ans[i][3]);
61     return 0;
62 }
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posted on 2016-02-13 22:57  onlyRP  阅读(271)  评论(0编辑  收藏  举报