HDU-1264 Counting Squares 矩形面积并 线段树+扫描线

题意：

     给任意个矩形求矩形的并集

 

交集算的面积为覆盖一次或一次以上的面积

并集为覆盖两次或两次以上的面积 代码就多个求覆盖两次或两次的函数

可以参考矩形面积交

http://blog.csdn.net/u011742541/article/details/13233973

 

#include "stdio.h"
#include "algorithm"
using namespace std;
const int maxn = 110;
int n,pos;
int y[maxn*2];
struct node
{
int x,y1,y2;
int f;
}Line[maxn*2];

struct Node
{
int y1,y2,len,inlen;
int ld,rd,c;
}tree[maxn*4];

bool cmp( node a,node b )  
{  
   return a.x < b.x;  
} 
void swap( int &a,int &b )
{
int temp = a;
a = b;
b = temp;
}

void buildtree( int ld,int rd,int t )
{
tree[t].c = 0;
tree[t].y1 = y[ld]; tree[t].y2 = y[rd]; 
tree[t].ld = ld;    tree[t].rd = rd;
tree[t].len = 0;    tree[t].inlen = 0;
if( ld + 1 == rd )
return;
int mid = ( ld+rd )>>1;
buildtree( ld,mid,t<<1 );  
buildtree( mid,rd,t<<1|1 );  
}
void calen( int t )             //求线段树被覆盖1次长度   
{  
    if( tree[t].c > 0 )   // t区间f是否被覆盖   
    {  
        tree[t].len = tree[t].y2 - tree[t].y1;  
        return;  
    }  
    if( tree[t].ld +1 == tree[t].rd ) tree[t].len = 0;      
    else  
        tree[t].len = tree[t<<1].len + tree[t<<1|1].len;    //  回带len的值   
}  

void incalen(int t)          //求交集时用到 线段被覆盖两次或两次以上   
{  
    if(tree[t].c >= 2)   
    {  
        tree[t].inlen = tree[t].y2 - tree[t].y1;  
        return;  
    }  
    if( tree[t].ld+1 == tree[t].rd )  tree[t].inlen = 0;  
    else if( tree[t].c == 1 )   
    {  
        tree[t].inlen = tree[t<<1].len + tree[t<<1|1].len;  
    }  
    else tree[t].inlen = tree[t<<1].inlen + tree[t<<1|1].inlen;  
} 

void updata( int t,node e )
{
if( e.y1 == tree[t].y1 && e.y2 == tree[t].y2 )
{
tree[t].c += e.f;
calen(t);  
//incalen(t);//求交集时用到
return;
}
if( e.y2 <= tree[t<<1].y2 )
updata( t<<1,e ); 
else if( e.y1 >= tree[t<<1|1].y1 )  //e在tree[t]的右半边   
        updata( t<<1|1,e );  
else
{
node temp = e;
temp.y2 = tree[t<<1].y2;
updata( t<<1,temp ); 

temp = e;
temp.y1 = tree[t<<1|1].y1;
updata( t<<1|1,temp ); 
}
calen(t);  
    //incalen(t);//求交集时用到
}

int main()
{
//freopen( "data.txt","r",stdin );
    int x1,y1,x2,y2;
while( true )
{
pos = 0;
while( scanf("%d%d%d%d",&x1,&y1,&x2,&y2) == 4 && x1 != -1 && x1 != -2 )
{
if( x1 > x2 )               //把坐标1作为左上角 2作为右下角
swap( x1,x2 );
if( y1 > y2 )
swap( y1,y2 );
pos++;
Line[pos].x = x1;       Line[pos].f = 1;
Line[pos].y1 = y1;Line[pos].y2 = y2;
y[pos] = y1;

pos++;
Line[pos].x = x2;       Line[pos].f = -1;
Line[pos].y1 = y1;Line[pos].y2 = y2;
y[pos] = y2;

}
sort( y+1,y+pos+1 );
sort( Line+1,Line+pos+1,cmp );
buildtree( 1,pos,1 );
int ans = 0;
for( int i = 1; i <= pos; i ++ )
{
ans += tree[1].len * ( Line[i].x - Line[i-1].x ); 
updata( 1,Line[i] );  
}
printf("%d\n",ans);
if( x1 == -2 )
break;
}
    return 0;
}