Picture

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.

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.

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.

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

228

IOI 1998

矩形周长并,  和面积并差不多, 每新插入一条边,周长的增量就是上一次总长和这一次总长的差值的绝对值, 我是先算横边,再算纵边的, 在纸上画一下图就知道了

/*************************************************************************
    > File Name: POJ1177.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2015年01月15日 星期四 15时13分56秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 20110;

struct node
{
	int l, r;
	int len;
	int add;
}tree[N << 2];

struct seg
{
	int l, r;
	int h;
	int flag;
}lines[N << 1], lines2[N << 1];

int cmp (seg a, seg b)
{
	if (a.h != b.h)
	{
		return a.h < b.h;
	}
	return a.flag > b.flag;
}

void build (int p, int l, int r)
{
	tree[p].l = l;
	tree[p].r = r;
	tree[p].len = 0;
	tree[p].add = 0;
	if (l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	build (p << 1, l, mid);
	build (p << 1 | 1, mid + 1, r);
}

void pushup (int p)
{
	if (tree[p].add)
	{
		tree[p].len = tree[p].r - tree[p].l + 1;
	}
	else if (tree[p].l == tree[p].r)
	{
		tree[p].len = 0;
	}
	else
	{
		tree[p].len = tree[p << 1].len + tree[p << 1 | 1].len;
	}
}

void update (int p, int l, int r, int val)
{
	if (l == tree[p].l && r == tree[p].r)
	{
		tree[p].add += val;
		pushup (p);
		return;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (r <= mid)
	{
		update (p << 1, l, r, val);
	}
	else if (l > mid)
	{
		update (p << 1 | 1, l, r, val);
	}
	else
	{
		update (p << 1, l, mid, val);
		update (p << 1 | 1, mid + 1, r, val);
	}
	pushup (p);
}

int main()
{
	int n;
	int x1, y1, x2, y2;
	while (~scanf("%d", &n))
	{
		build (1, 1, N);
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
			x1 += 10001;
			x2 += 10001;
			lines[i].flag = 1;
			lines[i].l = x1;
			lines[i].r = x2;
			lines[i].h = y1;
			lines[i + n].flag = -1;
			lines[i + n].l = x1;
			lines[i + n].r = x2;
			lines[i + n].h = y2;
			y1 += 10001;
			y2 += 10001;
			lines2[i].flag = 1;
			lines2[i].l = y1;
			lines2[i].r = y2;
			lines2[i].h = x1;
			lines2[i + n].flag = -1;
			lines2[i + n].l = y1;
			lines2[i + n].r = y2;
			lines2[i + n].h = x2;
		}
		sort (lines + 1, lines + 2 * n + 1, cmp);
		sort (lines2 + 1, lines2 + 2 * n + 1, cmp);
		int  ans = 0;
		for (int i = 1; i <= 2 * n; ++i)
		{
			int last = tree[1].len;
			update (1, lines[i].l, lines[i].r - 1, lines[i].flag);
			ans += abs(last - tree[1].len);
		}
		build (1, 1, N);
		for (int i = 1; i <= 2 * n; ++i)
		{
			int last = tree[1].len;
			update (1, lines2[i].l, lines2[i].r - 1, lines2[i].flag);
			ans += abs(last - tree[1].len);
		}
		printf("%d\n", ans);
	}
	return 0;
}