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.
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.
InputYour 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.OutputYour 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 16Sample Output
228
题意 ： 求一堆矩形的周长并，覆盖的的地方周长不计算
思路 ： 类似于矩形的周长并 ， 只不过不算面积 ， 同时线段树中要多维护一个竖边的个数 。
怎么求 图示的红线部分的变呢 ？ 就是当前的边的长度 - 上一次边的长度。 好好理解下 。
还有一个很重要的地方是 就是判断两条边是否重叠或相交 好好理解下子 。
代码示例 ：
/*
* Author: ry
* Created Time: 2017/10/24 11:11:20
* File Name: 1.cpp
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <time.h>
using namespace std;
const int eps = 1e4+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long
struct seg
{
int l, r, h;
int f;
}po[eps];
int x[eps];
bool cmp(seg a, seg b){
return a.h < b.h;
}
struct node
{
int l, r, len, f;
int lf, rf, num;
}tree[eps<<2];
void build(int l, int r, int k){
tree[k].l = l;
tree[k].r = r;
tree[k].f = 0;
tree[k].len = tree[k].num = 0;
tree[k].lf = tree[k].rf = 0;
if (l == r) return;
int m = (l + r) >> 1;
build(l, m, k<<1);
build(m+1, r, k<<1|1);
}
void down(int k){
if (tree[k].f) {
tree[k].len = x[tree[k].r+1] - x[tree[k].l];
tree[k].lf = tree[k].rf = 1;
tree[k].num = 1;
}
else if (tree[k].l == tree[k].r){
tree[k].len = 0;
tree[k].lf = tree[k].rf = 0;
tree[k].num = 0;
}
else {
tree[k].len = tree[k<<1].len + tree[k<<1|1].len;
tree[k].lf = tree[k<<1].lf;
tree[k].rf = tree[k<<1|1].rf;
tree[k].num = tree[k<<1].num + tree[k<<1|1].num - (tree[k<<1].rf&&tree[k<<1|1].lf);
}
}
void update(int l, int r, int k, int pt){
if (l <= tree[k].l && tree[k].r <= r){
tree[k].f += pt;
down(k);
return;
}
int m = (tree[k].l + tree[k].r) >> 1;
if (l <= m) update(l, r, k<<1, pt);
if (r > m) update(l, r, k<<1|1, pt);
down(k);
}
int main() {
int n;
int a, b, c, d;
while (~scanf("%d", &n)){
int k = 1;
for(int i = 0; i < n; i++){
scanf("%d%d%d%d", &a, &b, &c, &d);
po[k].l = po[k+1].l = a;
po[k].r = po[k+1].r = c;
po[k].h = b, po[k+1].h = d;
po[k].f = 1, po[k+1].f = -1;
x[k] = a, x[k+1] = c;
k += 2;
}
sort(x+1, x+k);
sort(po+1, po+k, cmp);
int t = 2;
for(int i = 2; i < k; i++){
if (x[i] != x[i-1]) x[t++] = x[i];
}
build(1, t-1, 1);
int ans = 0, last = 0;
po[k].h = po[k-1].h
; for(int i = 1; i < k; i++){
int l = lower_bound(x+1, x+t, po[i].l) - x;
int r = lower_bound(x+1, x+t, po[i].r) - x - 1;
update(l, r, 1, po[i].f);
ans += 2*tree[1].num*(po[i+1].h - po[i].h);
ans += fabs((tree[1].len - last));
last = tree[1].len;
}
printf("%d
", ans);
}
return 0;
}