覆盖的面积
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3553 Accepted Submission(s): 1743
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.
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
#include<iostream>
#include<cstdio>
#include<algorithm>
#define Max 10005
using namespace std;
struct line{
double x, y1, y2;
int flag;
}x_line[Max]; struct node{
int l, r, flag;
double x, f;
}tree[Max]; double point[Max];
int n, m, xm; int cmp(double a,double b)
{
return a<b;
} bool comp(line a,line b)
{
return a.x<b.x;
} void Build(int l,int r,int k)
{
int m;
tree[k].l = l;
tree[k].r = r;
tree[k].flag = 0;
tree[k].x = 0.0;
tree[k].f = 0.0;
if(l+1 == r) return;
m = (l+r)>>1;
Build(l,m,k+k);
Build(m,r,k+k+1);
} void Myscanf()
{
m = 1;
double x1,x2,y1,y2;
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
point[m] = y1;
x_line[m].x = x1;
x_line[m].y1 = y1;
x_line[m].flag = 1;
x_line[m++].y2 = y2;
point[m] = y2;
x_line[m].x = x2;
x_line[m].y1 = y1;
x_line[m].flag = -1;
x_line[m++].y2 = y2;
} } void Point_Do()
{
int mi = 0;
sort(point+1,point+m,cmp);
sort(x_line+1,x_line+m,comp);
xm = m;
for(int i=1; i<m; i++)
{
if(point[i]!=point[mi]) point[++mi] = point[i]; //去重
}
m = mi;
} int Bin(double xi)
{
int l = 1, r = m;
int mi;
while(l<=r)
{
mi = (l+r)>>1;
if(xi == point[mi]) return mi;
if(xi > point[mi]) l = mi+1;
else r = mi-1;
}
return -1;
} void update(int l, int r, int k, line cur)
{
if(tree[k].l==tree[k].r-1)
{
if(tree[k].flag == 1 && cur.flag == 1) tree[k].f = cur.x;
else if(tree[k].flag==2 && cur.flag==-1){
tree[k].x +=( cur.x - tree[k].f );
tree[k].f = 0.0;
}
tree[k].flag+=cur.flag;
return;
}
int mi = (tree[k].l+tree[k].r)>>1;
if(l>=mi) update(l,r,k+k+1,cur);
else if(r<=mi) update(l,r,k+k,cur);
else{
update(l,mi,k+k,cur);
update(mi,r,k+k+1,cur);
}
return;
} double query(int k)
{
if(tree[k].l + 1==tree[k].r) return (point[tree[k].r]-point[tree[k].l])*tree[k].x;
return query(k+k) + query(k+k+1);
} int Ans()
{
for(int i=1;i<xm;i++)
{
int x = Bin(x_line[i].y1);
int y = Bin(x_line[i].y2);
update(x,y,1,x_line[i]);
}
} int main()
{
int Case;
scanf("%d",&Case);
while(Case--)
{
memset(tree,0,sizeof(tree));
memset(point,-1,sizeof(point));
memset(x_line,0,sizeof(x_line));
Myscanf();
Point_Do();
Build(1,m,1);
Ans();
printf("%.2lf\n",query(1));
} return 0;
}
主要思路就是将y轴离散为线段树，然后，将x轴转换为状态，其实也可以x为线段树，y为状态的
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