HDU-1255 覆盖的面积 线段树 + 扫描线

http://acm.hdu.edu.cn/showproblem.php?pid=1225

求矩形面积交 主要是扫描线不太懂 纠结了很久

竖向建线段树 横向扫描线

以扫描线扫描出相邻2条竖边的距离 再以线段树求出两竖向边的交集面积高度

tree[1].inlen * ( Line[i].x - Line[i-1].x ); tree[1].len 表示f为1的线段树长度 Line[i].x - Line[i-1].x相邻2边的距离

#include "stdio.h"
#include "algorithm"
using namespace std;
const int maxn = 2010;
int n;
double y[maxn];
struct node
{
double x,y1,y2;    //x 边的位置  y边的区间大小
int f;                //f 标记 前边还是后边
}Line[maxn];
struct node1
{
double y1,y2,len,inlen;    //  inlen线段被覆盖两次的长度
int ld,rd,c;                 // c为被覆盖次数
}tree[maxn*4];

bool cmp( node a,node b )
{
return a.x < b.x;
}

void buildtree( int ld,int rd,int t)  // 建树
{
tree[t].c = 0;      tree[t].inlen = 0;
tree[t].ld = ld;     tree[t].rd = rd;
tree[t].y1 = y[ld]; tree[t].y2 = y[rd];
tree[t].len = 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)          //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.y2 < tree[t].y1 || e.y1 > tree[t].y2 )
return;
if( e.y1 == tree[t].y1 && e.y2 == tree[t].y2 )   //在线段树中找到边e所在的区间 更新f求出长度冷
{
tree[t].c += e.f;
calen(t);
incalen(t);
return;
}
if( e.y2 <= tree[t<<1].y2 )   //如果e全在tree[t]的左半边
updata( t<<1,e );
else if( e.y1 >= tree[t<<1|1].y1 )  //e在tree[t]的右半边
updata( 2*t+1,e );
else if( e.y2 > tree[t<<1|1].y1 && e.y1 < tree[t<<1].y2 )                               //一部分在左半边 一部分在右半边
{
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);
}
int main()
{
//freopen("data.txt","r",stdin);
int i,j,cas,pos;
double x1,y1,x2,y2,ans;
scanf("%d",&cas);
while( cas-- )
{
scanf("%d",&n);
pos = 0;
for( i = 1; i <= n; i ++ )
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

pos++;
Line[pos].x = x1;                 //存前边
Line[pos].y1 = y1;
Line[pos].y2 = y2;
Line[pos].f = 1;
y[pos] = y1;

pos++;
Line[pos].x = x2;                  //存后边
Line[pos].y1 = y1;
Line[pos].y2 = y2;
Line[pos].f = -1;
y[pos] = y2;
}
sort( y+1,y+pos+1);			  //对横边进行排序
sort( Line+1,Line+pos+1,cmp );          //对竖边进行排序
buildtree( 1,pos,1 );
ans = 0;
for( i = 1; i <= pos; i ++ )   //当某一条线段被覆盖两次或两次以上 计算一次面积
{
ans += tree[1].inlen * ( Line[i].x - Line[i-1].x );  //tree[1].len 表示f为1的线段树长度 Line[i].x - Line[i-1].x相邻2边的距离
updata( 1,Line[i] );
}
printf("%.2lf\n",ans);
}
return 0;
}