HDU 1828 扫描线（矩形周长并）

Picture

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3310    Accepted Submission(s): 1723

Problem Description

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.

 

Input

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.

Please process to the end of file.

 

Output

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.

 

Sample Input

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

 

Sample Output

228

 

Source

IOI 1998

 

题目意思：

给n个矩形，求并后的总周长

 

思路：

矩形周长并，很经典的扫描线题目。周长=水平线+竖直线。

求水平线的长度时，每插入一条线段后总长度减去插入线段之前的总长度即为增长的长度，最终即得出水平线的长度。

求竖直线的长度时，每插入一条线段后“裸露”出的点的个数乘上后一线段高度和这一线段的高度差，即得出竖直线长度。

 

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 5005
#define ll root<<1
#define rr root<<1|1
#define mid (a[root].l+a[root].r)/2

int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<0?-x:x;}

struct node{
    int l, r;
    int sum;
    int ysum;
    int val;
    bool lv, rv;
}a[N*8];

struct Line{
    int x1, x2, y;
    int val;
    Line(){}
    Line(int a,int b,int c,int d){
        x1=a;
        x2=b;
        y=c;
        val=d;
    }
}line[N*2];

bool cmp(Line a,Line b){
    return a.y<b.y;
}
int n, m;
int xx[N*2];

int b_s(int key){
    int l=1, r=m;
    while(l<=r){
        int mm=(l+r)/2;
        if(xx[mm]==key) return mm;
        else if(xx[mm]>key) r=mm-1;
        else if(xx[mm]<key) l=mm+1;
    }
}

void build(int l,int r,int root){
    a[root].l=l;
    a[root].r=r;
    a[root].ysum=a[root].sum=a[root].val=0;
    a[root].lv=a[root].rv=false;
    if(l==r) return;
    build(l,mid,ll);
    build(mid+1,r,rr);
}

void up(int root){
    if(a[root].val){
        a[root].sum=xx[a[root].r+1]-xx[a[root].l];
        a[root].lv=a[root].rv=true;
        a[root].ysum=2;    
    } 
    else if(a[root].l==a[root].r) {
        a[root].lv=a[root].rv=false;
        a[root].ysum=a[root].sum=0;
    }
    else{
        a[root].sum=a[ll].sum+a[rr].sum;
        a[root].lv=a[ll].lv;
        a[root].rv=a[rr].rv;
        a[root].ysum=a[ll].ysum+a[rr].ysum;
        if(a[rr].lv&&a[ll].rv) a[root].ysum-=2;
    } 
}

void update(int l,int r,int val,int root){
    if(a[root].l==l&&a[root].r==r){
        a[root].val+=val;
        up(root);
        return;
    }
    if(l>=a[rr].l) update(l,r,val,rr);
    else if(r<=a[ll].r) update(l,r,val,ll);
    else{
        update(l,mid,val,ll);
        update(mid+1,r,val,rr);
    }
    up(root);
}

main()
{
    int i, j, k;
    int x1, y1, x2, y2;
    int kase=1;
    while(scanf("%d",&n)==1&&n){
        m=1;k=0;
        for(i=0;i<n;i++){
            scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
            line[k++]=Line(x1,x2,y1,1);
            line[k++]=Line(x1,x2,y2,-1);
            xx[m++]=x1;xx[m++]=x2;
        }
        sort(xx+1,xx+m);
        m=unique(xx+1,xx+m)-xx-1;
        sort(line,line+k,cmp);
        build(1,m,1);
        int ans=0;
        int pre=0;
        for(i=0;i<k-1;i++){
            update(b_s(line[i].x1),b_s(line[i].x2)-1,line[i].val,1);
            ans+=abs(a[1].sum-pre)+(line[i+1].y-line[i].y)*a[1].ysum;
            pre=a[1].sum;
        }
        update(b_s(line[k-1].x1),b_s(line[k-1].x2)-1,line[k-1].val,1);
        ans+=abs(a[1].sum-pre);
        printf("%d
",ans);
    }
}