覆盖的面积
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3870 Accepted Submission(s): 1906
Problem Description
给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.
Input
输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐标的范围从0到100000.
注意:本题的输入数据较多,推荐使用scanf读入数据.
Output
对于每组测试数据,请计算出被这些矩形覆盖过至少两次的区域的面积.结果保留两位小数.
Sample Input
2 5 1 1 4 2 1 3 3 7 2 1.5 5 4.5 3.5 1.25 7.5 4 6 3 10 7 3 0 0 1 1 1 0 2 1 2 0 3 1
Sample Output
7.63 0.00
Author
Ignatius.L & weigang Lee
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这题用线段树维护时,要多维护一个len2,表示被覆盖至少2次的线段长度, 其他和矩形并一样, 在纸上画一下图就知道为什么了
当add >= 2
区间里被覆盖一次及一次以上的线段长度和被覆盖至少2次的线段长度一样,都是区间长度
当add==1
区间里被覆盖一次及一次以上的线段长度是区间长度, 被覆盖至少2次的线段长度是左右子树被覆盖至少一次的线段长度之和(这些线段在子树上被覆盖一次, 在根上又被覆盖一,所以一共被覆盖2次)
当add == 0
根状态由子树决定
注意特判叶子节点
/*************************************************************************
> File Name: hdu1255.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年01月15日 星期四 17时46分07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 = 2010;
struct node
{
double l, r;
double h;
int flag;
}lines[N << 1];
struct segment
{
int l, r;
int add;
double len1, len2;
}tree[N << 2];
double xis[N << 1];
int n, cnt;
int cmp (node a, node b)
{
return a.h < b.h;
}
int BinSearch(double val)
{
int l = 1;
int r = cnt;
int mid;
while (l <= r)
{
mid = (l + r) >> 1;
if (xis[mid] < val)
{
l = mid + 1;
}
else if (xis[mid] > val)
{
r = mid - 1;
}
else
{
break;
}
}
return mid;
}
void build (int p, int l, int r)
{
tree[p].l = l;
tree[p].r = r;
tree[p].add = 0;
tree[p].len1 = 0;
tree[p].len2 = 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 >= 2)
{
tree[p].len2 = xis[tree[p].r + 1] - xis[tree[p].l];
tree[p].len1 = tree[p].len2;
}
else if (tree[p].add == 1)
{
tree[p].len1 = xis[tree[p].r + 1] - xis[tree[p].l];
if (tree[p].l == tree[p].r)
{
tree[p].len2 = 0;
}
else
{
tree[p].len2 = tree[p << 1].len1 + tree[p << 1 | 1].len1;
}
}
else
{
if (tree[p].l == tree[p].r)
{
tree[p].len1 = tree[p].len2 = 0;
}
else
{
tree[p].len1 = tree[p << 1].len1 + tree[p << 1 | 1].len1;
tree[p].len2 = tree[p << 1].len2 + tree[p << 1 | 1].len2;
}
}
}
void update (int p, int l, int r, int val)
{
if (tree[p].l == 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 t;
double x1, y1, x2, y2;
scanf("%d", &t);
while (t--)
{
cnt = 0;
scanf("%d", &n);
for (int i = 1; i <= n; ++i)
{
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
lines[i].l = x1;
lines[i].r = x2;
lines[i].h = y1;
lines[i].flag = 1;
lines[i + n].l = x1;
lines[i + n].r = x2;
lines[i + n].h = y2;
lines[i + n].flag = -1;
xis[++cnt] = x1;
xis[++cnt] = x2;
}
sort (xis + 1, xis + cnt + 1);
cnt = unique (xis + 1, xis + cnt + 1) - xis - 1;
build (1, 1, cnt);
sort (lines + 1, lines + 2 * n + 1, cmp);
double ans = 0;
int l = BinSearch (lines[1].l);
int r = BinSearch (lines[1].r) - 1;
update (1, l, r, lines[1].flag);
for (int i = 2; i <= 2 * n; ++i)
{
ans += tree[1].len2 * (lines[i].h - lines[i - 1].h);
l = BinSearch (lines[i].l);
r = BinSearch (lines[i].r) - 1;
update (1, l, r, lines[i].flag);
}
printf("%.2lf\n", ans);
}
return 0;
}