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Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3075    Accepted Submission(s): 1616

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

给出 n 个矩阵 求包围这些矩阵的边界长度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
using namespace std;
typedef pair<int,int> pii ;
typedef long long LL;
#define X first
#define Y second
#define root 1,n,1
#define lr rt<<1
#define rr rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int N = 200010;
const int M = 10000;
const int mod = 10007;
int n , m ;
struct Point { int x , y ;Point(){} };
struct Line{ int tag ; Point a, b ; }e[N];
inline bool cmp1 ( const Line &A , const Line &B ) { return A.a.x < B.a.x ; }
inline bool cmp2 ( const Line &A , const Line &B ) { return A.a.y < B.a.y ; }

int lazy[N<<2] , cnt[N<<2] , sum[N<<2] ;

void build( int l , int r , int rt ) {
    sum[rt] = lazy[rt] = cnt[rt] = 0 ;
    if( l == r ) return ;
    int mid = (l+r)>>1;
    build(lson),build(rson);
}

void Up( int l , int r , int rt ) {
    if( cnt[rt] > 0 ) sum[rt] = r - l + 1 ;
    else sum[rt] = sum[lr] + sum[rr] ;
}

void update( int l , int r , int rt , int L , int R , int tag ) {
    if( l == L && r == R ) {
        if( tag ) {
            cnt[rt]++ , lazy[rt] ++ ;
            sum[rt] = r - l + 1 ;
        }
        else {
            cnt[rt]-- , lazy[rt] -- ;
            if( cnt[rt] > 0 ) sum[rt] = r - l + 1 ;
            else {
                if( l == r ) sum[rt] = 0 ;
                else sum[rt] = sum[lr] + sum[rr] ;
            }
        }
        return ;
    }
    int mid = (l+r)>>1;
    if( L > mid ) update(rson,L,R,tag);
    else if( R <= mid ) update(lson,L,R,tag);
    else update(lson,L,mid,tag) , update(rson,mid+1,R,tag);
    Up(l,r,rt);
}

int x1[N] , x2[N] , y1[N] , y2[N];

int main()
{
    #ifdef LOCAL
       freopen("in.txt","r",stdin);
//       freopen("out.txt","w",stdout);
    #endif // LOCAL
    int _ , cas = 1 ;
    int mx , Mx , my  , My ;
    while( scanf("%d",&n) != EOF ) {
        Mx = My = -N , mx = my = N ;
        for( int i = 0 ; i < n ; ++i ) {
            scanf("%d%d%d%d",&x1[i],&y1[i],&x2[i],&y2[i]);
            mx = min( mx , x1[i] ); Mx = max( Mx , x2[i] );
            my = min( my , y1[i] ); My = max( My , y2[i] );
            e[i].a.x = x1[i] , e[i].a.y = y1[i]  ;
            e[i].b.x = x1[i] , e[i].b.y = y2[i] ;
            e[i].tag = 1 ;
            e[i+n].a.x = x2[i] , e[i+n].a.y = y2[i];
            e[i+n].b.x = x2[i] , e[i+n].b.y = y1[i];
            e[i+n].tag = 0 ;
        }
        int tot = n * 2 ;
        sort( e , e + tot ,cmp1 ) ;
        LL ans = 0 , last = 0 ;
        build( my , My - 1 , 1 );
        for( int i = 0 ; i < tot ; ++i ) {
            int x = e[i].a.y , y = e[i].b.y ;
            if( x > y ) swap(x,y);
            update( my , My - 1 , 1 , x , y -1 , e[i].tag );
            LL tmp = sum[1] ;
            ans += abs( tmp - last );
            last = tmp ;
        }
        for( int i = 0 ; i < n ; ++i ){
            e[i].a.x = x1[i] , e[i].a.y = y1[i]  ;
            e[i].b.x = x2[i] , e[i].b.y = y1[i] ;
            e[i].tag = 1 ;
            e[i+n].a.x = x2[i] , e[i+n].a.y = y2[i]  ;
            e[i+n].b.x = x1[i] , e[i+n].b.y = y2[i] ;
            e[i+n].tag = 0 ;
        }
        last = 0 ;
        sort( e , e + tot , cmp2 ) ;
        build(mx,Mx-1,1);
        for( int i = 0 ; i < tot ; ++i ) {
            int x = e[i].a.x , y = e[i].b.x ;
            if( x > y ) swap(x,y);
            update( mx , Mx - 1 , 1 , x , y - 1 , e[i].tag );
            LL tmp = sum[1] ;
            ans += abs( tmp - last );
            last = tmp ;
        }
        printf("%I64d\n",ans);
    }
}