POJ 1389 Area of Simple Polygons(线段树+扫描线求矩形面积并)

//9379418 NKHelloWorld 1389 Accepted 296K 16MS C++ 1997B 2011-10-01 12:31:16
//9379419 NKHelloWorld 1389 Accepted 844K 0MS G++ 1997B 2011-10-01 12:31:31
/*
POJ1151的整数简化版本，套用模板1A
*/
#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;

#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )

int y[3000];

struct Line
{
	int x, y1, y2,flag;
} line[3000];

struct Node
{
	int l, r, cover;
	int lf, rf, len;
} node[10000];
bool cmp ( Line a, Line b )
{
	return a.x < b.x;
}

void length ( int u )
{
	if ( node[u].cover > 0 )
	{
		node[u].len = node[u].rf - node[u].lf;
		return;
	}
	else if ( node[u].l + 1 == node[u].r )
		node[u].len = 0; /* 叶子节点，len 为 0 */
	else
	    node[u].len = node[L(u)].len + node[R(u)].len;
}

void build ( int u, int l, int r )
{
	node[u].l = l; node[u].r = r;
	node[u].lf = y[l]; node[u].rf = y[r];
	node[u].len = node[u].cover = 0;
	if ( l + 1 == r ) return;
	int mid = ( l + r ) / 2;
	build ( L(u), l, mid );
	build ( R(u), mid, r );
}

void update ( int u, Line e )
{
	if ( e.y1 == node[u].lf && e.y2 == node[u].rf )
	{
		node[u].cover += e.flag;
		length ( u );
		return;
	}
	if ( e.y1 >= node[R(u)].lf )
		update ( R(u), e );
	else if ( e.y2 <= node[L(u)].rf )
		update ( L(u), e );
	else
	{
		Line temp = e;
		temp.y2 = node[L(u)].rf;
		update ( L(u), temp );
		temp = e;
		temp.y1 = node[R(u)].lf;
		update ( R(u), temp );
	}
	length ( u );
}

int main()
{
	int n, t,i;
	int x1,y1,x2,y2,ans;
	while(1)
	{
	    t = n = 1;
	    while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2),x1!=-1)
	    {
	        line[t].x = x1; line[t].y1 = y1;    line[t].y2 = y2;    line[t].flag = 1;   y[t] = y1;  t++;
	        line[t].x = x2; line[t].y1 = y1;    line[t].y2 = y2;    line[t].flag = -1;  y[t] = y2;  t++;
	        n++;
	    }
	    if(n==1)    break;
	    n--;
		sort ( line + 1, line + t, cmp );
		sort ( y + 1, y + t );
		build ( 1, 1, t-1 );
        update ( 1, line[1] );
		ans = 0;
		for ( i = 2; i < t; i++ )
		{
			ans += node[1].len * ( line[i].x - line[i-1].x );
			update ( 1, line[i] );
		}
		printf("%d\n",ans);
	}
	return 0;
}