hdu 1542 Atlantis線分樹面積を求め、、、ニマデータは本当に穴があいていて、配列は決して小さくしてはいけません!
Atlantis
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7815 Accepted Submission(s): 3420
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<=x1The 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
Sample Output
面積を求めてそして:
おすすめブログ:http://www.cnblogs.com/ka200812/archive/2011/11/13/2247064.html
マイコード:
君と励まし合う
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7815 Accepted Submission(s): 3420
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
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
面積を求めてそして:
おすすめブログ:http://www.cnblogs.com/ka200812/archive/2011/11/13/2247064.html
マイコード:
#include <cstdio>
#include <algorithm>
using namespace std ;
double y[300] ;
struct Line{
double x , y_up , y_down ;
int mark ;
}line[300] ; // 110,WA ！！ discuss A !
struct Node{
double x ,y_up,y_down ;
int cover ;
bool isLeaf ;
}st[400100];
bool cmp(const Line &a , const Line &b)
{
return a.x<b.x ;
}
void build(int l ,int r , int pos)
{
st[pos].cover = 0 ;
st[pos].y_down = y[l] ;
st[pos].y_up = y[r] ;
st[pos].x = -1 ;
st[pos].isLeaf = false ;
if(l+1 == r)
{
st[pos].isLeaf = true ;
return ;
}
int mid = (l+r)>>1 ;
build(l,mid,pos<<1);
build(mid,r,pos<<1|1) ;
}
double insert(double x , double y_up , double y_down , int mark , int pos)
{
if(st[pos].y_down>=y_up || st[pos].y_up<=y_down)
{
return 0 ;
}
if(st[pos].isLeaf)
{
if(st[pos].cover>0)
{
double temp = st[pos].x ;
double area = (x-temp)*(st[pos].y_up-st[pos].y_down) ;
st[pos].x = x ;
st[pos].cover += mark ;
return area ;
}
else
{
st[pos].x = x ;
st[pos].cover += mark ;
return 0 ;
}
}
return insert(x,y_up,y_down,mark,pos<<1)+insert(x,y_up,y_down,mark,pos<<1|1) ;
}
int main()
{
int n , c = 1;
while(~scanf("%d",&n) && n)
{
int index = 0 ;
for(int i = 0 ; i < n ; ++i)
{
double x1,y1,x2,y2 ;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2) ;
line[index].x = x1 ;
line[index].y_down = y1 ;
line[index].y_up = y2 ;
y[index] = y1 ;
line[index++].mark = 1 ;
line[index].x = x2 ;
line[index].y_down = y1 ;
line[index].y_up = y2 ;
y[index] = y2 ;
line[index++].mark = -1 ;
}
sort(y,y+index) ;
sort(line,line+index,cmp) ;
build(0,index-1,1) ;
double area = 0.0 ;
for(int i = 0 ; i < index ; ++i)
{
area += insert(line[i].x,line[i].y_up,line[i].y_down,line[i].mark,1) ;
}
printf("Test case #%d
",c++) ;
printf("Total explored area: %.2lf
",area) ;
}
return 0 ;
}君と励まし合う