POJ 1389 (discretization + computational geometry)

Area of simple polygons

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3412		Accepted: 1763

Description

There is n, 1 <= n <= rectangles in the 2-d xy-plane. The four sides of a rectangle is horizontal or vertical line segments. Rectangles is defined by their lower-left and Upper-right corner points. Each corner point was a pair of nonnegative integers in the range of 0 through 50,000 indicating its x and Y coordinate S.

Assume that the contour of their union was defi Ned by a set S of segments. We can use a subset of S to construct simple polygon (s). Please report the total area of the polygon (s) constructed by the subset of S. The area should is as large as possible. In a 2-d xy-plane, a polygon are defined by a finite set of segments such this every segment extreme (or endpoint) is share D by exactly edges and no subsets of edges have the same property. The segments is edges and their extremes are the vertices of the polygon. A polygon is simple if there are no pair of nonconsecutive edges sharing a point.

Example:consider the following three rectangles:

Rectangle 1: < (0, 0) (4, 4);

Rectangle 2: < (1, 1) (5, 2);

Rectangle 3: < (1, 1) (2, 5);.

The polygons constructed by these rectangles is 18.

Input

The input consists of multiple test cases. A line of 4-1 ' s separates each test case. An extra line of 4-1 ' s marks the end of the input. In each test case, the rectangles is given one by one in a line. In each line for a rectangle, 4 non-negative integers is given. The first is the x and Y coordinates of the lower-left corner. The next is the x and Y coordinates of the upper-right corner.

Output

For each test case, output the total area of all simple polygons in a line.

Sample Input

Sample Output

Test instructions: Rectangular area intersection.
Only the discretization of the useless line tree is also 47MS AC. It's a bit of a lot of space, but opening a vis array into a bool type should be much less space.

///discretization of#include <iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<cmath>using namespacestd;Const intN =2005;structrec{intX1,y1,x2,y2;} Rec[n];intX[n],y[n];intVis[n][n];intk,t;intBinary1 (intvalue) {    intMid,l=0, r=k-1;  while(l<R) {Mid= (l+r) >>1; if(X[mid]==value)returnmid; if(x[mid]<value) L = mid+1; Elser = mid-1; }    returnl;}intBinary2 (intvalue) {    intMid,l=0, r=k-1;  while(l<S) {Mid= (l+r) >>1; if(Y[mid]==value)returnmid; if(y[mid]<value) L = mid+1; Elser = mid-1; }    returnl;}voidinput () {intX1,y1,x2,y2;  while(true) {scanf ("%d%d%d%d",&x1,&y1,&x2,&y2); if(x1==-1&&x2==-1&&y1==-1&&y2==-1) Break; Rec[t].x1= X1,rec[t].y1 = Y1,rec[t].x2=x2,rec[t++].y2 =Y2; X[K]= x1,y[k++] =Y1; X[K]= x2,y[k++] =Y2; } sort (X,x+k); Sort (Y,y+k);}voidsolve () {intT1,t2,t3,t4;  for(intI=0; i<t;i++) {T1=Binary1 (REC[I].X1); T2=Binary1 (REC[I].X2); T3=Binary2 (rec[i].y1); T4=Binary2 (rec[i].y2);  for(intj=t1;j<t2;j++){             for(intL = t3;l<t4;l++) {Vis[j][l]=1; }        }    }    intArea =0;  for(intI=0; i<k;i++){         for(intj=0; j<k;j++) { area+=vis[i][j]* (x[i+1]-x[i]) * (y[j+1]-Y[j]); }} printf ("%d\n", area);}intMain () {intX1,y1,x2,y2;  while(SCANF ("%d%d%d%d", &x1,&y1,&x2,&y2)! =EOF) {        if(x1==-1&&x2==-1&&y1==-1&&y2==-1) Break; memset (Vis,0,sizeof(VIS)); K=0, t=0; Rec[t].x1= X1,rec[t].y1 = Y1,rec[t].x2=x2,rec[t++].y2 =Y2; X[K]= x1,y[k++] =Y1; X[K]= x2,y[k++] =Y2;        Input ();    Solve (); }    return 0;}

POJ 1389 (discretization + computational geometry)