bzoj 1818: [Cqoi2010]内部白点

Posted 2020-06-23 xiyuedong

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 struct data
 6 {
 7     int x,y,k,r;
 8 }a[1000008];
 9 struct dat
10 {
11     int x,y;
12 }b[1000008];
13 bool cmp(dat a1,dat a2)
14 {
15     if(a1.x==a2.x)
16       return a1.y<a2.y;
17     return a1.x<a2.x;
18 }
19 int n,h[100008],h1[100009],shu[1000008],cnt,ans;
20 void jia1(int a1,int a2)
21 {
22     cnt++;
23     a[cnt].x=lower_bound(h+1,h+h[0]+1,b[a1].x)-h;
24     a[cnt].y=b[a1].y;
25     a[cnt].k=1;
26     cnt++;
27     a[cnt].x=a[cnt-1].x;
28     a[cnt].y=b[a2].y;
29     a[cnt].k=-1;
30 }
31 bool cmp1(dat a1,dat a2)
32 {
33     if(a1.y==a2.y)
34       return a1.x<a2.x;
35     return a1.y<a2.y;
36 }
37 void jia2(int a1,int a2)
38 {
39     cnt++;
40     a[cnt].y=b[a1].y;   
41     a[cnt].x=lower_bound(h+1,h+h[0]+1,b[a1].x)-h;
42     a[cnt].r=lower_bound(h+1,h+h[0]+1,b[a2].x)-h;
43 }
44 bool cmp2(data a1,data a2)
45 {
46     if(a1.y==a2.y)
47       return a1.k<a2.k;
48     return a1.y<a2.y;
49 }
50 int zhao(int a1)
51 {
52     int sum=0;
53     for(;a1;a1-=a1&-a1)
54       sum+=shu[a1];
55     return sum;
56 }
57 void gai(int a1,int a2)
58 {
59     for(;a1<=h[0];a1+=a1&-a1)
60       shu[a1]+=a2;
61     return;
62 }
63 int main()
64 {
65     scanf("%d",&n);
66     for(int i=1;i<=n;i++)
67       scanf("%d%d",&b[i].x,&b[i].y);
68     for(int i=1;i<=n;i++)
69       h1[i]=b[i].x;
70     sort(h1+1,h1+1+n);
71     h[1]=h1[1];
72     h[0]=1;
73     for(int i=2;i<=n;i++)
74       if(h1[i]!=h1[i-1])
75         {
76             h[0]++;
77             h[h[0]]=h1[i];
78         }
79     sort(b+1,b+n+1,cmp);
80     for(int i=2;i<=n;i++)
81       if(b[i].x==b[i-1].x)
82         jia1(i-1,i);
83     sort(b+1,b+n+1,cmp1);
84     for(int i=2;i<=n;i++)
85       if(b[i].y==b[i-1].y)
86         jia2(i-1,i);
87     sort(a+1,a+cnt+1,cmp2);
88     for(int i=1;i<=cnt;i++)
89       if(a[i].k==0)
90         ans+=zhao(a[i].r-1)-zhao(a[i].x);
91       else
92         gai(a[i].x,a[i].k);
93     printf("%d",ans+n);
94     return 0;
95 }

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 struct data
 6 {
 7     int x,y,k,r;
 8 }a[1000008];
 9 struct dat
10 {
11     int x,y;
12 }b[1000008];
13 bool cmp(dat a1,dat a2)
14 {
15     if(a1.x==a2.x)
16       return a1.y<a2.y;
17     return a1.x<a2.x;
18 }
19 int n,h[100008],h1[100009],shu[1000008],cnt,ans;
20 void jia1(int a1,int a2)
21 {
22     cnt++;
23     a[cnt].x=lower_bound(h+1,h+h[0]+1,b[a1].x)-h;
24     a[cnt].y=b[a1].y;
25     a[cnt].k=1;
26     cnt++;
27     a[cnt].x=a[cnt-1].x;
28     a[cnt].y=b[a2].y;
29     a[cnt].k=-1;
30 }
31 bool cmp1(dat a1,dat a2)
32 {
33     if(a1.y==a2.y)
34       return a1.x<a2.x;
35     return a1.y<a2.y;
36 }
37 void jia2(int a1,int a2)
38 {
39     cnt++;
40     a[cnt].y=b[a1].y;   
41     a[cnt].x=lower_bound(h+1,h+h[0]+1,b[a1].x)-h;
42     a[cnt].r=lower_bound(h+1,h+h[0]+1,b[a2].x)-h;
43 }
44 bool cmp2(data a1,data a2)
45 {
46     if(a1.y==a2.y)
47       return a1.k<a2.k;
48     return a1.y<a2.y;
49 }
50 int zhao(int a1)
51 {
52     int sum=0;
53     for(;a1;a1-=a1&-a1)
54       sum+=shu[a1];
55     return sum;
56 }
57 void gai(int a1,int a2)
58 {
59     for(;a1<=h[0];a1+=a1&-a1)
60       shu[a1]+=a2;
61     return;
62 }
63 int main()
64 {
65     scanf("%d",&n);
66     for(int i=1;i<=n;i++)
67       scanf("%d%d",&b[i].x,&b[i].y);
68     for(int i=1;i<=n;i++)
69       h1[i]=b[i].x;
70     sort(h1+1,h1+1+n);
71     h[1]=h1[1];
72     h[0]=1;
73     for(int i=2;i<=n;i++)
74       if(h1[i]!=h1[i-1])
75         {
76             h[0]++;
77             h[h[0]]=h1[i];
78         }
79     sort(b+1,b+n+1,cmp);
80     for(int i=2;i<=n;i++)
81       if(b[i].x==b[i-1].x)
82         jia1(i-1,i);
83     sort(b+1,b+n+1,cmp1);
84     for(int i=2;i<=n;i++)
85       if(b[i].y==b[i-1].y)
86         jia2(i-1,i);
87     sort(a+1,a+cnt+1,cmp2);
88     for(int i=1;i<=cnt;i++)
89       if(a[i].k==0)
90         ans+=zhao(a[i].r-1)-zhao(a[i].x);
91       else
92         gai(a[i].x,a[i].k);
93     printf("%d",ans+n);
94     return 0;
95 }

将点拆成横线与点，用树状数组维护。