Poj 1177 picture (scanning line for perimeter)

It is similar to finding the correct area, but it is not well handled when the Scanning direction is the same. If the scanning line is discretization twice, this problem can be solved, however, this is a waste of time and space. Here, because I am a discrete X coordinate, it is difficult to calculate statistics parallel to the Y axis. The solution is to mark the breakpoint on the left of the interval and the breakpoint on the right to find the total number of breakpoints in the interval. You can calculate the length parallel to the Y axis. When merging, You need to determine whether the left and right intervals on the right are the same. If they are the same, it means they are connected together, minus more.

Picture

Time limit:2000 ms		Memory limit:10000 K
Total submissions:10300		Accepted:5462

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 [-keys, 10000] and any existing rectangle has a positive area.

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 2515 -4 24 140 -6 16 42 15 10 2230 10 36 2034 0 40 16

Sample output

228

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <ctime>#include <map>#include <set>#define eps 1e-12///#define M 1000100#define LL __int64///#define LL long long///#define INF 0x7ffffff#define INF 0x3f3f3f3f#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 50100;struct node{    int l, r;    int h;    int x;} f[maxn];int dc[maxn];int cnt[maxn<<2];int sum[maxn<<2];int ans[maxn<<2];int Left[maxn<<2];int Right[maxn<<2];bool cmp(node a, node b){    return a.h < b.h;}int Find(int x, int a[], int n){    int l = 0;    int r = n-1;    while(l <= r)    {        int mid = (l+r)>>1;        if(a[mid] == x) return mid;        else if(a[mid] < x) l = mid+1;        else r = mid-1;    }    return -1;}void Up(int l, int r, int site){    if(cnt[site])    {        sum[site] = dc[r+1]-dc[l];        Left[site] = Right[site] = 1;        ans[site] = 2;    }    else if(l == r)    {        sum[site] = 0;        ans[site] = 0;        Left[site] = Right[site] = 0;    }    else    {        sum[site] = sum[site<<1]+sum[site<<1|1];        ans[site] = ans[site<<1]+ans[site<<1|1];        Left[site] = Left[site<<1];        Right[site] = Right[site<<1|1];        if(Left[site<<1|1] && Right[site<<1]) ans[site] -= 2;    }}void Update(int l, int r, int L, int R, int d, int site){    if(L <= l && r <= R)    {        cnt[site] += d;        Up(l, r, site);        return;    }    int mid = (l+r)>>1;    if(L <= mid) Update(l, mid, L, R, d, site<<1);    if(R > mid) Update(mid+1, r, L, R, d, site<<1|1);    Up(l, r, site);}int main(){    int n;    while(cin >>n)    {        if(!n) break;        int x1, y1, x2, y2;        int m = 0;        for(int i = 0; i < n; i++)        {            scanf("%d %d %d %d",&x1, &y1, &x2, &y2);            dc[m] = x1;            f[m].l = x1;            f[m].r = x2;            f[m].h = y1;            f[m++].x = 1;            dc[m] = x2;            f[m].l = x1;            f[m].r = x2;            f[m].h = y2;            f[m++].x = -1;        }        sort(dc, dc+m);        sort(f, f+m, cmp);        int k = unique(dc, dc+m)-dc;        memset(cnt, 0, sizeof(cnt));        memset(sum, 0, sizeof(sum));        memset(ans, 0, sizeof(ans));        memset(Left, 0, sizeof(Left));        memset(Right, 0, sizeof(Right));        int res = 0;        int x = 0;        for(int i = 0; i < m; i++)        {            int l = Find(f[i].l, dc, k);            int r = Find(f[i].r, dc, k)-1;            if(l <= r) Update(0, k-1, l, r, f[i].x, 1);            if(i+1 < m) res += ans[1]*(f[i+1].h-f[i].h);            res += abs(sum[1]-x);            x = sum[1];        }        cout<<res<<endl;    }    return 0;}