POJ 1151 Atlantis 线段树+扫描线

解题思路:

先将y轴进行离散化。n个矩形的2n个横边纵坐标共构成最多2n-1个区间的边界，对这些区间编号，建立起线段树。

x轴记录左边和右边，左边时是矩形面积增加，覆盖层数增加边，右边是形面积减少，覆盖层数减少边。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 using namespace std;
  5 double y[210];
  6 int ncount;
  7 struct line
  8 {
  9     double x,y1,y2;
 10     bool left;
 11 };
 12 line lines[210];
 13 struct node
 14 {
 15     int l,r;
 16     node *pl,*pr;
 17     double len;
 18     int c;
 19 };
 20 bool cmp(const line &a,const line &b)
 21 {
 22     return a.x<b.x;
 23 }
 24 node t[1010];
 25 void build(node *root,int l,int r)
 26 {
 27     root->l=l;
 28     root->r=r;
 29     root->c=0;
 30     root->len=0;
 31     if(l==r) return;
 32     ncount++;
 33     root->pl=t+ncount;
 34     ncount++;
 35     root->pr=t+ncount;
 36     build(root->pl,l,(l+r)/2);
 37     build(root->pr,(l+r)/2+1,r);
 38 }
 39 int searchy(int e,double x)
 40 {
 41     int s=0;
 42     e=e-1;
 43     while(s<=e)
 44     {
 45         int mid=s+(e-s)/2;
 46         if(y[mid]==x)
 47             return mid;
 48         else if(y[mid]>x)
 49             e=mid-1;
 50         else s=mid+1;
 51     }
 52     return -1;
 53 }
 54 int mid(node *p)
 55 {
 56     return (p->l+p->r)/2;
 57 }
 58 void Insert(node *root,int l,int r)
 59 {
 60     if(root->l==l&&root->r==r)
 61     {
 62         root->len=y[r+1]-y[l];
 63         root->c++;
 64         return ;
 65     }
 66     if(r<=mid(root))
 67         Insert(root->pl,l,r);
 68     else if(l>=mid(root)+1)
 69         Insert(root->pr,l,r);
 70     else
 71     {
 72         Insert(root->pl,l,mid(root));
 73         Insert(root->pr,mid(root)+1,r);
 74     }
 75     if(root->c==0)
 76         root->len=root->pl->len+root->pr->len;
 77 }
 78 void Delete(node *root,int l,int r)
 79 {
 80     if(root->l==l&&root->r==r)
 81     {
 82         root->c--;
 83         if(root->c==0)
 84         {
 85             if(root->l==root->r)
 86                 root->len=0;
 87             else
 88                 root->len=root->pl->len+root->pr->len;
 89         }
 90         return;
 91     }
 92     if(r<=mid(root))
 93         Delete(root->pl,l,r);
 94     else if(l>=mid(root)+1)
 95         Delete(root->pr,l,r);
 96     else
 97     {
 98         Delete(root->pl,l,mid(root));
 99         Delete(root->pr,mid(root)+1,r);
100     }
101     if(root->c==0)
102         root->len=root->pl->len+root->pr->len;
103 }
104 int main()
105 {
106     int n;
107     double x1,x2,y1,y2;
108     int yc,lc;
109     int ca=0;
110     while(~scanf("%d",&n)&&n)
111     {
112         ca++;
113         lc=yc=0;
114         for(int i=0; i<n; i++)
115         {
116             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
117             y[yc++]=y1;
118             y[yc++]=y2;
119             lines[lc].x=x1;
120             lines[lc].y1=y1;
121             lines[lc].y2=y2;
122             lines[lc].left=1;
123             lc++;
124 
125             lines[lc].x=x2;
126             lines[lc].y1=y1;
127             lines[lc].y2=y2;
128             lines[lc].left=0;
129             lc++;
130         }
131         sort(y,y+yc);
132         yc=unique(y,y+yc)-y;
133         build(t,0,yc-2);
134         sort(lines,lines+lc,cmp);
135         double ans=0;
136         for(int i=0; i<lc-1; i++)
137         {
138             int l=searchy(yc,lines[i].y1);
139             int r=searchy(yc,lines[i].y2)-1;
140             if(lines[i].left)
141                 Insert(t,l,r);
142             else
143                 Delete(t,l,r);
144             ans+=t[0].len*(lines[i+1].x-lines[i].x);
145         }
146         printf("Test case #%d\n",ca);
147         printf("Total explored area: %.2f\n\n",ans);
148     }
149 }