2013
11-09

# Picture

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of all rectangles is called the perimeter.

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1.

The corresponding boundary is the whole set of line segments drawn in Figure 2.

The vertices of all rectangles have integer coordinates.

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate.

0 <= number of rectangles < 5000

All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

228

import java.io.*;

public class Main{

static A4 p[];
static A5 ST[];

public static void main(String[] args) throws Exception{

p = new A4[2*n+1];
ST = new A5[40005];
int j = 0;
int min=10001,max=-10001;

for(int i=0;i< n;i++){
int x1 = Integer.parseInt(str[0]);
int y1 = Integer.parseInt(str[1]);
int x2 = Integer.parseInt(str[2]);
int y2 = Integer.parseInt(str[3]);
p[j++] = new A4(x1,y1,y2,1);
p[j++] = new A4(x2,y1,y2,-1);
min = Math.min(y1, min);
max = Math.max(y2, max);
}

java.util.Arrays.sort(p,0,j);
build(1,min,max);
cal(j);

}

public static void cal(int n) {

int sum = 0,t = 0;
p[n] = p[n-1];   // 最后一个没有 平行x轴的长度

for(int i=0;i< n;i++){
insert(1,p[i].y1,p[i].y2,p[i].s);
sum += ST[1].time*(p[i+1].x-p[i].x)*2;// 加上平行x轴的长度
sum += Math.abs(ST[1].len-t);   //减去被覆盖的长度
t = ST[1].len;   // t的值为   当前 被覆盖的 平行 y轴的长度
}

System.out.println(sum);

}

public static void insert(int k,int l, int r, int s) {

if(l<=ST[k].l&&r>=ST[k].r){
ST[k].count += s;
update(k);
return ;
}

int mid = (ST[k].l+ST[k].r)>>1;
if(l< mid)
insert(2*k,l,r,s);
if(r>mid)
insert(2*k+1,l,r,s);
update(k);

}

public static void update(int k) {

if(ST[k].count>0){
ST[k].len = ST[k].r-ST[k].l;  // 覆盖的线段长度
ST[k].lp=ST[k].rp=ST[k].time=1; // 因为完全覆盖了此区间，所以lp,rp,time = 1;
}else if(ST[k].r - ST[k].l == 1){   // 避免 越界
ST[k].len = ST[k].lp = ST[k].rp = ST[k].time = 0;
}else{
int l = 2*k;int r = 2*k+1;
ST[k].len = ST[l].len + ST[r].len; //总覆盖长度
//如果 左子树右端点被覆盖且右子树左端点也被覆盖刚此是连续的，所以减1
ST[k].time = ST[l].time + ST[r].time - (ST[l].rp&ST[r].lp);

ST[k].lp = ST[l].lp;
ST[k].rp = ST[r].rp;
}

}

public static void build(int k, int s, int t) {  //建树

ST[k] = new A5(s,t,0,0,0,0,0);

if(t - s <= 1)
return ;

int mid = (s+t)>>1;
build(2*k,s,mid);
build(2*k+1,mid,t);

}

}

class A4 implements Comparable< A4>{
int x;  //扫描线 x 坐标
int y1;
int y2;
int s;
public A4(int x, int y1, int y2, int s) {
super();
this.x = x;
this.y1 = y1;
this.y2 = y2;
this.s = s;
}

public int compareTo(A4 e) {

if(this.x< e.x)
return -1;
else if(this.x>e.x)
return 1;
else{
if(this.s>e.s)
return -1;
else if(this.s< e.s)
return 1;
else
return 0;
}
}
}
class A5{
int l,r,count,len,lp,rp,time;

public A5(int l, int s, int count, int len, int lp, int rp, int time) {
super();
this.l = l;
this.r = s;
this.count = count;
this.len = len;
this.lp = lp;
this.rp = rp;
this.time = time;
}

}

1. 很高兴你会喜欢这个网站。目前还没有一个开发团队，网站是我一个人在维护，都是用的开源系统，也没有太多需要开发的部分，主要是内容整理。非常感谢你的关注。

2. 这道题目的核心一句话是：取还是不取。
如果当前取，则index+1作为参数。如果当前不取，则任用index作为参数。