poj 1389(离散化+计算几何)

Area of Simple Polygons

Description

There are N, 1 <= N <= 1,000 rectangles in the 2-D xy-plane. The four sides of a rectangle are horizontal or vertical line segments. Rectangles are defined by their lower-left and upper-right corner points. Each corner point is a pair of two nonnegative integers in the range of 0 through 50,000 indicating its x and y coordinates.

Assume that the contour of their union is 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 be as large as possible. In a 2-D xy-plane, a
polygon is defined by a finite set of segments such that every segment
extreme (or endpoint) is shared by exactly two edges and no subsets of
edges has the same property. The segments are edges and their extremes
are the vertices of the polygon. A polygon is simple if there is 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 total area of all simple 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 are given one by one in a line. In each line
for a rectangle, 4 non-negative integers are given. The first two are
the x and y coordinates of the lower-left corner. The next two are 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

0 0 4 4
1 1 5 2
1 1 2 5
-1 -1 -1 -1
0 0 2 2
1 1 3 3
2 2 4 4
-1 -1 -1 -1
-1 -1 -1 -1　 

Sample Output

18
10 

题意：矩形面积交.只用了离散化没用线段树也 47MS AC。。就是耗费的空间大了点，，不过将vis数组开成bool类型应该可以少很多空间..

///离散化
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
const int N = 2005;
struct Rec{
    int x1,y1,x2,y2;
}rec[N];
int x[N],y[N];
int vis[N][N];
int k,t;
int binary1(int value){
    int mid,l=0,r=k-1;
    while(l<r){
        mid = (l+r)>>1;
        if(x[mid]==value) return mid;
        if(x[mid]<value) l = mid+1;
        else r = mid-1;
    }
    return l;
}
int binary2(int value){
    int mid,l=0,r=k-1;
    while(l<r){
        mid = (l+r)>>1;
        if(y[mid]==value) return mid;
        if(y[mid]<value) l = mid+1;
        else r = mid-1;
    }
    return l;
}
void input(){
    int x1,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);
}
void solve(){
    int t1,t2,t3,t4;
    for(int i=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(int j=t1;j<t2;j++){
            for(int l = t3;l<t4;l++){
                vis[j][l] = 1;
            }
        }
    }
    int area = 0;
    for(int i=0;i<k;i++){
        for(int j=0;j<k;j++){
            area+=vis[i][j]*(x[i+1]-x[i])*(y[j+1]-y[j]);
        }
    }
    printf("%d\n",area);
}
int main()
{
    int x1,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;
}