HDU covers an area of 1255 square meters (rectangular area and multiple times)

Covered Area

Time Limit: 10000/5000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 1949 accepted submission (s): 960


Problem description is a number of rectangles on the plane. The area of the area covered by these rectangles is obtained at least twice.

 

The first line of input data is a positive integer T (1 <= T <= 100), representing the number of test data. the first row of each test data is a positive integer N (1 <= n <= 1000), representing the number of rectangles, followed by N rows of data, each row contains four floating point numbers, it indicates the coordinates of the upper left corner and lower right corner of a rectangle on the plane. The upper and lower sides of the rectangle are parallel to the X axis, and the left and right sides are parallel to the Y axis. the coordinate ranges from 0 to 100000.

Note: This question contains a large amount of input data. We recommend that you use scanf to read data.

Output for each group of test data, calculate the area of the area covered by these rectangles at least twice. The result is retained with two decimal places.

Sample Input

251 1 4 21 3 3 72 1.5 5 4.53.5 1.25 7.5 46 3 10 730 0 1 11 0 2 12 0 3 1

 

Sample output

7.630.00

 

Authorignatius. L & Weigang Lee

Recommendignatius. L

Question: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1255

This is the area of the image that covers more than two times.

Analysis, but I did not expect a better method...

Code:

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#define ls rt<<1#define rs rt<<1|1#define lson l,m,ls#define rson m,r,rsusing namespace std;const int mm=2222;const int mn=mm<<2;struct seg{    double x,y1,y2;    int val;}g[mm];double len[mn],ret[mn],y[mm],L,R;int t[mn],val;void build(int n){    while(n--)ret[n]=len[n]=t[n]=0;}void updata(int l,int r,int rt){    if(L<=y[l]&&R>=y[r])t[rt]+=val;    else    {        int m=(l+r)>>1;        if(L<y[m])updata(lson);        if(R>y[m])updata(rson);    }    if(t[rt]>1)ret[rt]=len[rt]=y[r]-y[l];    else if(t[rt]>0)len[rt]=y[r]-y[l],ret[rt]=len[ls]+len[rs];    else if(l>=r)ret[rt]=len[rt]=0;    else len[rt]=len[ls]+len[rs],ret[rt]=ret[ls]+ret[rs];}bool cmp(seg a,seg b){    return a.x<b.x;}int main(){    int i,m,n,cs;    double ans;    scanf("%d",&cs);    while(cs--)    {        scanf("%d",&n);        for(i=0;i<n;++i)        {            scanf("%lf%lf%lf%lf",&g[i].x,&y[i],&g[i+n].x,&y[i+n]);            g[i].y1=y[i],g[i].y2=y[i+n],g[i].val=1;            g[i+n].y1=y[i],g[i+n].y2=y[i+n],g[i+n].val=-1;        }        sort(y,y+n+n);        sort(g,g+n+n,cmp);        for(m=i=0;i<n+n;++i)            if(y[m]<y[i])y[++m]=y[i];        build(m<<2);        for(ans=i=0;i<n+n;++i)        {            L=g[i].y1,R=g[i].y2,val=g[i].val;            updata(0,m,1);            if(g[i].x<g[i+1].x)ans+=(g[i+1].x-g[i].x)*ret[1];        }        printf("%.2lf\n",ans);    }    return 0;}