1 //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include<cstdio>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<cmath>
  7 #define clr(x) memset(x,0,sizeof(x))
  8 #define MAXN 10010
  9 using namespace std;
 10 struct edg
 11 {
 12     double l,r,y;
 13     int d;
 14 }edge[MAXN];
 15 struct seg
 16 {
 17     int l,r,cov;
 18     double len1,len2;
 19 }segt[MAXN<<2];
 20 int cnt;
 21 double x[MAXN];
 22 bool cmp(edg a,edg b)
 23 {
 24     if(a.y==b.y) return a.d>b.d;
 25     return a.y<b.y;
 26 }
 27 double max(double a,double b)
 28 {
 29     return a>b?a:b;
 30 }
 31 void init(int i,int l,int r)
 32 {
 33     segt[i]=(seg){l,r,0,0,0};
 34     if(l==r)
 35         return ;
 36     int mid=(l+r)>>1;
 37     init(i<<1,l,mid);
 38     init(i<<1|1,mid+1,r);
 39     return ;
 40 }
 41 void pushup(int i)
 42 {
 43     if(segt[i].cov>=2)
 44     {
 45         segt[i].len2=segt[i].len1=x[segt[i].r+1]-x[segt[i].l];
 46     }
 47     else if(segt[i].cov==1)
 48     {
 49         segt[i].len1=x[segt[i].r+1]-x[segt[i].l];
 50         if(segt[i].l==segt[i].r)
 51             segt[i].len2=0;
 52         else
 53             segt[i].len2=max(segt[i<<1].len1,segt[i<<1].len2)+max(segt[i<<1|1].len1,segt[i<<1|1].len2);
 54     }
 55     else
 56     {
 57         if(segt[i].l==segt[i].r)
 58         {
 59             segt[i].len1=segt[i].len2=0;
 60         }
 61         else
 62         {
 63             segt[i].len2=segt[i<<1].len2+segt[i<<1|1].len2;
 64             segt[i].len1=segt[i<<1].len1+segt[i<<1|1].len1;
 65         }
 66     }
 67     return ;
 68 }
 69 void update(int i,int l,int r,int value)
 70 {
 71     if(segt[i].l>=l && segt[i].r<=r)
 72     {
 73         segt[i].cov+=value;
 74         pushup(i);
 75         return ;
 76     }
 77     int mid=(segt[i].l+segt[i].r)>>1;
 78     if(mid>=r)
 79     {
 80         update(i<<1,l,r,value);
 81     }
 82     else if(mid<l)
 83     {
 84         update(i<<1|1,l,r,value);
 85     }
 86     else
 87     {
 88         update(i<<1,l,r,value);
 89         update(i<<1|1,l,r,value);
 90     }
 91     pushup(i);
 92     return ;
 93 }
 94 int main()
 95 {
 96     int T,n,m,k,u,v;
 97     double x1,x2,y1,y2,ans,l,r;
 98     scanf("%d",&T);
 99     while(T--)
100     {
101         scanf("%d",&n);
102         cnt=0;
103         ans=0;
104         for(int i=1;i<=n;i++)
105         {
106             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
107             edge[++cnt]=(edg){x1,x2,y1,1};
108             x[cnt]=x1;
109             edge[++cnt]=(edg){x1,x2,y2,-1};
110             x[cnt]=x2;
111         }
112         n<<=1;
113         sort(x+1,x+n+1);
114         m=unique(x+1,x+n+1)-x-1;
115         sort(edge+1,edge+n+1,cmp);
116         init(1,1,m);
117         for(int i=1;i<n;i++)
118             if(edge[i].r>edge[i].l)
119             {
120                 l=lower_bound(x+1,x+m+1,edge[i].l)-x;
121                 r=lower_bound(x+1,x+m+1,edge[i].r)-x;
122                 update(1,l,r-1,edge[i].d);
123                     ans+=segt[1].len2*(edge[i+1].y-edge[i].y);
124             }
125         printf("%0.2lf\n",ans);
126     }
127     return 0;
128 }

Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.

Output

For each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

Sample Input

2 10 10 20 20 15 15 25 25.5 0

Sample Output

Test case #1 Total explored area: 180.00

Source

Mid-Central European Regional Contest 2000

---

卡格式，不说了，都是泪。

  1 //#pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include<cstdio>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<algorithm>
  6 #include<cmath>
  7 #define clr(x) memset(x,0,sizeof(x))
  8 #define MAXN 10010
  9 using namespace std;
 10 struct edg
 11 {
 12     double l,r,y;
 13     int d;
 14 }edge[MAXN];
 15 struct seg
 16 {
 17     int l,r,cov;
 18     double len;
 19 }segt[MAXN<<2];
 20 int cnt;
 21 double x[MAXN];
 22 bool cmp(edg a,edg b)
 23 {
 24     if(a.y==b.y) return a.d>b.d;
 25     return a.y<b.y;
 26 }
 27 double max(double a,double b)
 28 {
 29     return a>b?a:b;
 30 }
 31 void init(int i,int l,int r)
 32 {
 33     segt[i]=(seg){l,r,0,0};
 34     if(l==r)
 35         return ;
 36     int mid=(l+r)>>1;
 37     init(i<<1,l,mid);
 38     init(i<<1|1,mid+1,r);
 39     return ;
 40 }
 41 void pushup(int i)
 42 {
 43     if(segt[i].cov)
 44     {
 45         segt[i].len=x[segt[i].r+1]-x[segt[i].l];
 46     }
 47     else if(segt[i].l==segt[i].r)
 48     {
 49         segt[i].len=0;
 50     }
 51     else
 52     {
 53         segt[i].len=segt[i<<1].len+segt[i<<1|1].len;
 54     }
 55     return ;
 56 }
 57 void update(int i,int l,int r,int value)
 58 {
 59     if(segt[i].l>=l && segt[i].r<=r)
 60     {
 61         segt[i].cov+=value;
 62         pushup(i);
 63         return ;
 64     }
 65     int mid=(segt[i].l+segt[i].r)>>1;
 66     if(mid>=r)
 67     {
 68         update(i<<1,l,r,value);
 69     }
 70     else if(mid<l)
 71     {
 72         update(i<<1|1,l,r,value);
 73     }
 74     else
 75     {
 76         update(i<<1,l,r,value);
 77         update(i<<1|1,l,r,value);
 78     }
 79     pushup(i);
 80     return ;
 81 }
 82 int main()
 83 {
 84     int T,n,m,k,u,v;
 85     double x1,x2,y1,y2,ans,l,r;
 86     int kase=0;
 87     while(scanf("%d",&n) && n!=0)
 88     {
 89         printf("Test case #%d\n",++kase);
 90         cnt=0;
 91         ans=0;
 92         for(int i=1;i<=n;i++)
 93         {
 94             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
 95             edge[++cnt]=(edg){x1,x2,y1,1};
 96             x[cnt]=x1;
 97             edge[++cnt]=(edg){x1,x2,y2,-1};
 98             x[cnt]=x2;
 99         }
100         n<<=1;
101         sort(x+1,x+n+1);
102         m=unique(x+1,x+n+1)-x-1;
103         sort(edge+1,edge+n+1,cmp);
104         init(1,1,m);
105         for(int i=1;i<n;i++)
106             if(edge[i].r>edge[i].l)
107             {
108                 l=lower_bound(x+1,x+m+1,edge[i].l)-x;
109                 r=lower_bound(x+1,x+m+1,edge[i].r)-x;
110                 update(1,l,r-1,edge[i].d);
111                     ans+=segt[1].len*(edge[i+1].y-edge[i].y);
112             }
113         printf("Total explored area: %0.2lf\n",ans);
114         printf("\n");
115     }
116     return 0;
117 }