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Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4529    Accepted Submission(s): 2237


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 線段樹，掃描線之周長並。 有兩種解法： 第一種：以橫縱軸各建線段樹， 並從橫縱座標各掃描一遍，掃描的時候求出長度差（sum【1】-未加入第i條邊的sum【1】）的絕對值，求出總和就是結果，和求面積並的時候差不多的過程。 第二種：以橫軸建線段樹，用上面的方法只能求出圖形的上下底邊長度，而左右邊的長度怎麼求？這就需要我們求出能分成多少不連續的線段，之後用（SS【i+1】.h - SS【i】.h）* numseg 求出左右邊長度。 程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <queue>
#define mem(p,k) memset(p,k,sizeof(p));
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x6fffffff
#define LL long long
#define MAXN 20000
using namespace std;
int col[MAXN<<2];
int sum[MAXN<<2],X[MAXN],lbd[MAXN<<2],rbd[MAXN<<2],numseg[MAXN<<2];
struct Seg{
    int l,r,h;
    int cur;
    Seg(){}
    Seg(int a,int b,int c,int d):l(a),r(b),h(c),cur(d){}
    bool operator <(const Seg &a)const{
        if(a.h==h)return cur>a.cur;
        return h<a.h;
   }
}ss[MAXN];
void pushup(int rt,int l,int r){
    if(col[rt]){
        sum[rt]=X[r+1]-X[l];
        lbd[rt]=rbd[rt]=1;
        numseg[rt]=2;
    }
    else if(l==r)sum[rt]=lbd[rt]=rbd[rt]=numseg[rt]=0;
    else {
            lbd[rt]=lbd[rt<<1];
            rbd[rt]=rbd[rt<<1|1];
            sum[rt]=sum[rt<<1]+sum[rt<<1|1];
            numseg[rt]=numseg[rt<<1]+numseg[rt<<1|1];
            if(rbd[rt<<1]&&lbd[rt<<1|1])numseg[rt]-=2;
    }
}

void update(int L,int R,int c,int l, int r,int rt){
    if(l>=L&&r<=R){
        col[rt]+=c;
        pushup(rt,l,r);
        return ;
    }
    int m=(l+r)>>1;
    if(L<=m)update(L,R,c,lson);
    if(R>m) update(L,R,c,rson);
    pushup(rt,l,r);
}
int main(){
    int n,k,pp=0;
    while(cin>>n){
        int ans=0,la=0;
        k=0;
        mem(sum,0);
        mem(col,0);
        mem(numseg,0);
        for(int i=0;i<n;i++){
            int a,b,c,d;
            scanf("%d%d%d%d",&a,&b,&c,&d);
            X[k]=a;
            ss[k++]=Seg(a,c,b,1);
            X[k]=c;
            ss[k++]=Seg(a,c,d,-1);
        }
        sort(ss,ss+k);
        sort(X,X+k);
        int kk=1;
        for(int i=1;i<k;i++){
            if(X[i]!=X[i-1])X[kk++]=X[i];
        }
        for(int i=0;i<k;i++){
            int l=lower_bound(X,X+kk,ss[i].l)-X;
            int r=lower_bound(X,X+kk,ss[i].r)-X-1;
            if(l<=r)update(l,r,ss[i].cur,0,kk-1,1);
            //cout<<numseg[1]<<"::"<<sum[1]<<endl;
            ans+=numseg[1]*(ss[i+1].h-ss[i].h);
            ans+=abs(sum[1]-la);
            la=sum[1];
        }
        printf("%d\n",ans);
    }
    return 0;
}