Language: Default Picture Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 11200   Accepted: 5915 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

[Submit]   [Go Back]   [Status]   [Discuss]

首先，周長是什麼？

周長是線段的長度。 是哪些線段呢？  投影下來的線段。  那麼投影下來的線段在原矩形的什麼位置呢。  投影下來的線段 一定是在原矩形的4邊上。

用2根 平行於x軸的直線進行掃描, 然後再沿著y軸方向進行投影。 如果一根線段（X1,X2,Y）被覆蓋了，那麼一定存在一邊yi < Y， 存在另外一邊yj > Y。所以自下往上，第一次投影下來（沿著y軸投影）的位置一定是周長， 然後第二次的如果落在了 第一次投影的線段中，那麼我們知道 重疊的不能計算。  那麼第二次的全部投影長度-第一次的全部投影長度就是新增的橫向的投影周長。 縱向的 一根的長度很好知道， 那麼有多少根呢？ 其實就是染色區間的數量。  坑點就是 區間要分三類討論。。。 

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_back
#define lson l,m,rt<<1
#define rson m,r,rt<<1|1

struct Line
{
	int l,r,y,s;
	Line() {};
	Line(int ll,int rr,int yy,int ss):l(ll),r(rr),y(yy),s(ss){};
}line[10010];
int X[10010];
int color[10010*4];
int numseg[10010*4];
int len[10010*4];
bool lbd[10010*4],rbd[10010*4];

int cmp(Line a,Line b)
{
	return a.y < b.y;
}

int n;
int t;

void build(int l,int r,int rt)
{
	lbd[rt] = rbd[rt] = 0;
	len[rt] = 0;
	numseg[rt] = 0;
	color[rt] = 0;
	if(r - l == 1) return ;

	int m = (l + r) >> 1;
	build(lson);
	build(rson);
}

void PushUp(int rt,int l,int r)
{
	if(color[rt] > 0)
	{
		lbd[rt] = rbd[rt] = true;
		len[rt] = X[r] - X[l];
		numseg[rt] = 1;
		return ;
	}
	if(r - l == 1)
	{
		lbd[rt] = rbd[rt] = false;
		len[rt] = 0;
		numseg[rt] = 0;
	}
	else
	{
		lbd[rt] = lbd[rt<<1];
		rbd[rt] = rbd[rt<<1|1];
		numseg[rt] = numseg[rt<<1] + numseg[rt<<1|1];
		if(rbd[rt<<1] && lbd[rt<<1|1])
		{
			numseg[rt] -= 1;
		}
		len[rt] = len[rt<<1] + len[rt<<1|1];
	}
}

void update(int L,int R,int add,int l,int r,int rt)
{
	if(L == X[l] && R == X[r])
	{
		color[rt] += add;
		PushUp(rt,l,r);
		return ;
	}

	int m = (l + r) >> 1;
	if(L <= X[m-1]) update(L,min(R,X[m]),add,lson);
	if(X[m] <= R-1) update(max(L,X[m]),R,add,rson);
	PushUp(rt,l,r);
}
int main()
{
	freopen("in.txt","r",stdin);
	while(scanf("%d",&n) != EOF)
	{
		t = 0;
		for(int i=0;i<n;i++)
		{
			int x1,y1,x2,y2;
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			line[t] = Line(x1,x2,y1,1);
			X[t ++] = x1;
			line[t] = Line(x1,x2,y2,-1);
			X[t ++] = x2;
		}
		sort(line,line+t,cmp);
		sort(X,X+t);
		int MAXN = unique(X,X+t) - X;
		build(0,MAXN-1,1); 
		int ans = 0;
		int pre = 0;
		for(int i=0;i<t-1;i++)
		{
			update(line[i].l,line[i].r,line[i].s,0,MAXN-1,1);
			ans += abs(len[1] - pre);
			pre = len[1];
			ans += numseg[1] * 2 * (line[i+1].y - line[i].y);
		}
		update(line[t-1].l,line[t-1].r,line[t-1].s,0,MAXN-1,1);
		ans += abs(len[1] - pre);
		printf("%d\n",ans);
	}
}