Farmer
John has N (1 <= N <= 25,000) rectangular barns on his farm, all
with sides parallel to the X and Y axes and integer corner coordinates
in the range 0..1,000,000. These barns do not overlap although they may
share corners and/or sides with other barns. Since he has extra cows to
milk this year, FJ would like to expand some of his barns. A barn has
room to expand if it does not share a corner or a wall with any other
barn. That is, FJ can expand a barn if all four of its walls can be
pushed outward by at least some amount without bumping into another
barn. If two barns meet at a corner, neither barn can expand. Please
determine how many barns have room to expand.
约翰有N(1≤N≤25000)个矩形牛棚，它们的墙均与坐标轴平行，而且其坐标在[o，1061范围内．任意两个牛棚不重叠，但可能会有公共的墙． 由于约翰的奶牛持续增加，他不得不考虑扩张牛棚．一个牛棚可以扩张，当且仅当它的四边均不与其它牛棚接触．如果两个牛棚有一个公共角，那它们均是不可扩张的．统计有多少牛棚可以扩张．
* Line 1: A single integer, N
* Lines 2..N+1: Four space-separated integers A, B, C, and D,
describing one barn. The lower-left corner of the barn is at (A,B) and
the upper right corner is at (C,D).
第1行输入N，之后N行每行输入一个牛棚的左下角和右上角坐标．
输出说明
* Line 1: A single integer that is the number of barns that can be expanded.
输出可扩张的牛棚数．
1 #include<set>
2 #include<cstdio>
3 #include<cstring>
4 #include<iostream>
5 #include<algorithm>
6 #define Rep(i,x,y) for(int i=x;i<y;++i)
7 #define For(i,x,y) for(int i=x;i<=y;++i)
8 using namespace std;
9 const int N = 3e5+5;
10 const int M = 1e6+5;
11 struct data{
12 int x,u,d,id;
13 bool kd;
14 bool operator < (const data&b) const{
15 return x<b.x||x==b.x&&kd<b.kd;
16 }
17 }a[N];
18 struct bit{
19 int c[M];
20 #define lowbit(x) (x&-x)
21 void insert(int x){
22 for(int i=x;i<M;i+=lowbit(i))
23 c[i]++;
24 }
25 void erase(int x){
26 for(int i=x;i<M;i+=lowbit(i))
27 c[i]--;
28 }
29 int query(int x){
30 if(!x) return 0;
31 int res=0;
32 for(int i=x;i;i-=lowbit(i))
33 res+=c[i];
34 return res;
35 }
36 void init(){
37 memset(c,0,sizeof(c));
38 }
39 }T1,T2;
40 int cnt;
41 int n;
42 bool nok[N];
43 int f[M][2];
44 void del(int i){
45 T1.erase(a[i].d);
46 T2.erase(a[i].u+1);
47 f[a[i].u][0]=0;
48 f[a[i].d][1]=0;
49 }
50 void INS(int i){
51 f[a[i].u][0]=a[i].id;
52 f[a[i].d][1]=a[i].id;
53 T1.insert(a[i].d);
54 T2.insert(a[i].u+1);
55 }
56 void ins(int l,int r){
57 For(i,l,r)
58 INS(i);
59 For(i,l,r){
60 del(i);
61 if(f[a[i].u][1]){
62 nok[a[i].id]=1;
63 nok[f[a[i].u][1]]=1;
64 }
65 if(f[a[i].d][0]){
66 nok[a[i].id]=1;
67 nok[f[a[i].d][0]]=1;
68 }
69 if(T1.query(a[i].u)-T2.query(a[i].d)>0) nok[a[i].id]=1;
70 INS(i);
71 }
72 }
73 int main(){
74 scanf("%d",&n);
75 For(i,1,n){
76 int l,r,u,d;
77 scanf("%d%d%d%d",&l,&d,&r,&u);
78 ++l;++r;++d;++u;
79 a[++cnt]=(data){l,u,d,i,0};
80 a[++cnt]=(data){r,u,d,i,1};
81 }
82 n<<=1;
83 sort(a+1,a+n+1);
84 for(int i=1;i<=n;){
85 int l=i;int r=i;
86 for(;r<=n&&a[r].x==a[l].x&&a[r].kd==0;++r);
87 ins(l,r-1);
88 for(;r<=n&&a[r].kd==1;++r)
89 del(r);
90 i=r;
91 }
92 memset(f,0,sizeof(f));
93 T1.init();
94 T2.init();
95 for(int i=n;i;){
96 int r=i;int l=i;
97 for(;l>0&&a[l].x==a[r].x&&a[l].kd==1;--l);
98 ins(l+1,r);
99 for(;l>0&&a[l].kd==0;--l) del(l);
100 i=l;
101 }
102 int res=0;
103 For(i,1,n/2) if(!nok[i]) ++res;
104 cout<<res<<endl;
105 return 0;
106 }
