覆盖的面积Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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
/*
hdu 1255 覆盖的面积(求覆盖至少两次以上的面积) 主要是在push_up那出现了问题
设定len是覆盖两次以上的长度，len1是覆盖1次以上的长度
那么有以下几种情况：
1.sum>1 则说当前区间被覆盖至少两次
2.l == r 直接为0
3.当前区间一次都没被覆盖过 lson.len + rson.len
4.当前区间被覆盖过一次sum=1,那么 lson.len1 + rson.len1，即左右孩子中被 //最开始没有考虑到这个情况TAT
至少覆盖过一次的和 hhh-2016-03-27 18:07:36
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
const int maxn = 50050;
struct node
{
int l,r,sum;
double len;
double len1;
int mid()
{
return (l+r)>>1;
}
} tree[maxn*5];
double hs[maxn];
int tot,m;
void push_up(int i)
{
if(tree[i].sum)
tree[i].len1 = hs[tree[i].r+1]-hs[tree[i].l];
else if(tree[i].l == tree[i].r)
tree[i].len1 = 0;
else
tree[i].len1 = tree[lson].len1+tree[rson].len1; if(tree[i].sum > 1)
tree[i].len = hs[tree[i].r+1]-hs[tree[i].l];
else if(tree[i].l == tree[i].r)
tree[i].len = 0;
else if(tree[i].sum == 1)
tree[i].len = tree[lson].len1 + tree[rson].len1;
else
tree[i].len = tree[lson].len+tree[rson].len;
} void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].sum=tree[i].len= 0;
tree[i].len1 = 0;
if(l ==r )
{
return ;
}
int mid=tree[i].mid();
build(lson,l,mid);
build(rson,mid+1,r);
push_up(i);
} void push_down(int i)
{ } void Insert(int i,int l,int r,int val)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].sum += val;
push_up(i);
return;
}
push_down(i);
int mid = tree[i].mid();
if(l <= mid)
Insert(lson,l,r,val);
if(r > mid)
Insert(rson,l,r,val);
push_up(i);
} struct edge
{
double l,r,high;
int va;
edge() {};
edge(double ll,double rr,double h,int v):l(ll),r(rr),high(h),va(v) {}
} tx[maxn]; bool cmp(edge a,edge b)
{
if(a.high != b.high)
return a.high < b.high;
else
return a.va > b.va;
} int fin(double x)
{
int l = 0,r = m-1;
while(l <= r)
{
int mid = (l+r)>>1;
if(hs[mid] == x)
return mid;
else if(hs[mid] < x)
l = mid+1;
else
r = mid-1;
}
} int main()
{
int n,x,q,t;
int cas =1;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
tot = 0;
double x1,x2,y1,y2;
for(int i = 1; i <= n; i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
hs[tot] = x1;
tx[tot++] = edge(x1,x2,y1,1);
hs[tot] = x2;
tx[tot++] = edge(x1,x2,y2,-1);
}
sort(hs,hs+tot);
sort(tx,tx+tot,cmp);
m = 1;
for(int i = 1; i < tot; i++)
if(hs[i] != hs[i-1])
hs[m++] = hs[i];
build(1,0,m-1);
double ans = 0;
for(int i = 0;i < tot;i++)
{
int l = fin(tx[i].l);
int r = fin(tx[i].r)-1;
//cout << tx[i].l <<" " << tx[i].r <<" "<<endl;
Insert(1,l,r,tx[i].va);
ans += tree[1].len*(tx[i+1].high-tx[i].high);
}
printf("%.2f\n",ans);
}
return 0;
}