POJ 1177 PICTURE

 

Picture

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 5093		Accepted: 2644

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.

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

 

/*

第一个线短树, 对于输入矩形，将其转换为所有竖边进行存储，然后将竖边进行排序

线段树的存储按照所有点y轴坐标递增序列的下标来组织

 

然后一次处理所有竖线，如果是矩形的左竖边，则将其加入线段树，如果是矩形的右竖边，则从线段树中减去

(注意，amountLen是指区域的覆盖长度，而不是总长度，即重复的区域只统计一次)，然后统计当前值，当前

值的统计分两块：（1）竖线长度 = 当前的amountLen - 上一个amountLen （2）横线长度 = （当前竖线x坐标 - 上一个竖线x坐标）× 当前的段总数 × 2； 最后 总长+= （1） + （2）即可

 

注意这道题的与star之类的题的不同点在于，star的amountLen是总长度，即包括重复区域，而这道题的总长度是覆盖长度，而不包括区域长度的重复统计

*/

#include <iostream> #include <algorithm> #define MAX_L 10000 using namespace std; //线短树结点 struct node { int l, r; //当前几点左右区间的索引(用来索引yIndex) node *ls, *rs; //左右子树 int len; //区间长度(即yIndex[r] - yIndex[l]) int segNum; //区间的个数 int cover; //当前区间被覆盖的次数 int amountLen; //被覆盖的区间的总长度 bool lcover, rcover; //左右端点是否被覆盖 node() { ls = rs = NULL; l = r = len = segNum = amountLen = cover = 0; lcover = rcover = false; } }; //存储所有的竖线，一个矩形会产生两条竖线 struct line{ int y1, y2; //竖线的上下断电的y坐标值 int x; //竖线的x坐标值 bool first; //记录这条竖线是矩形的第一条边否 bool operator < (const line &l) const { //根据x坐标排序时需要用到 return x < l.x; } }yLines[MAX_L]; int yIndex[MAX_L]; //存储唯一出现过的y坐标值 int recNum = 0; //输入的矩形个数 int uniqueYNum = 0; //独一无二的出现过的y值的个数 node *root; //建树，左右区间是[l, r] void makeTree(node *&root, int l, int r) { if(l < r){ root = new node(); root->l = l; root->r = r; root->len = yIndex[r] - yIndex[l]; if(r - l >= 2) { int mid = (l + r) / 2; makeTree(root->ls, l, mid); makeTree(root->rs, mid, r); } } } //更新当前节点被覆盖的总长度 void upDateAmountLen(node *&root) { //完全被覆盖了 if(root->cover > 0) root->amountLen = root->len; //cover = 0,未被覆盖且此节点是内部结点 else if(root->r - root->l >= 2) root->amountLen = root->ls->amountLen + root->rs->amountLen; //cover = 0,未被覆盖且是叶子结点 else root->amountLen = 0; } //更新当前节点被覆盖的线短的段数 void updateSegNum(node *&root) { //完全被覆盖了 if(root->cover > 0) { root->lcover = root->rcover = true; root->segNum = 1; //整个区间更成一个封闭的segment } //cover = 0,未被覆盖且此节点是内部结点 else if(root->r - root->l >= 2) { root->lcover = root->ls->lcover; root->rcover = root->rs->rcover; //如果左右是闭合的，则构成一整条线短需要减去1 root->segNum = root->ls->segNum + root->rs->segNum - root->ls->rcover * root->rs->lcover; } //cover = 0,未被覆盖且是叶子结点 else { root->lcover = root->rcover = false; root->segNum = 0; } } //插入或者删除一个区间,type = 0:插入;type = 1:删除 void modify(node *&root, int l, int r, int type) { if(root->l >= l && root->r <= r) { if(type == 0) root->cover++; else root->cover--; } else { int mid = (root->l + root->r) / 2; if(l < mid) modify(root->ls, l, r, type); if(r > mid) modify(root->rs, l, r, type); } upDateAmountLen(root); updateSegNum(root); } int getIndex(int y) { //return lower_bound(yIndex, yIndex + uniqueYNum, y) - yIndex; for(int i = 0; i < uniqueYNum; i++) if(yIndex[i] == y) return i; return -1; } int main() { int i, x1, y1, x2, y2; cin>>recNum; for(i = 0; i < recNum; i++) { cin>>x1>>y1>>x2>>y2; yLines[i * 2].x = x1; yLines[i * 2 + 1].x = x2; yLines[i * 2].y1 = yLines[i * 2 + 1].y1 = y1; yLines[i * 2].y2 = yLines[i * 2 + 1].y2 = y2; yLines[i * 2].first = true; yLines[i * 2 + 1].first = false; yIndex[i * 2] = y1; yIndex[i * 2 + 1] = y2; } //将所有出现过的y值进行排序 sort(yIndex, yIndex + recNum * 2); //对所有的竖线按照x坐标进行排序 sort(yLines, yLines + recNum * 2); //去除重复的index值 uniqueYNum = 0; for(i = 1; i < recNum * 2; i++) { if(yIndex[i - 1] != yIndex[i]) yIndex[uniqueYNum++] = yIndex[i - 1]; } yIndex[uniqueYNum++] = yIndex[2 * recNum - 1]; //建树 makeTree(root, 0, uniqueYNum - 1); int res = 0, lastLen = 0; //从左向右一次扫描 for(i = 0; i < 2 * recNum - 1; i++) { int type = 0; //如果是矩形的第一条树线则插入,若是第二条则删除 if(!yLines[i].first) type = 1; modify(root, getIndex(yLines[i].y1), getIndex(yLines[i].y2), type); //横线增加 res += root->segNum * (yLines[i + 1].x - yLines[i].x) * 2; //竖线增加 res += abs(root->amountLen - lastLen); lastLen = root->amountLen; } //处理最后一个并增加竖线 modify(root, getIndex(yLines[2 * recNum - 1].y1), getIndex(yLines[2 * recNum - 1].y2), 1); res += abs(root->amountLen - lastLen); cout<<res<<endl; return 0; }