POJ-1151-Atlantis（扫描线求面积并+线段树+离散化）

Atlantis

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11681		Accepted: 4537

Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

Input

The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area. 
The input file is terminated by a line containing a single 0. Don't process it.

Output

For each test case, your program should output one section. The first line of each section must be "Test case #k", where k is the number of the test case (starting with 1). The second one must be "Total explored area: a", where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point. 
Output a blank line after each test case.

Sample Input

2
10 10 20 20
15 15 25 25.5
0

Sample Output

Test case #1
Total explored area: 180.00 

Source

Mid-Central European Regional Contest 2000

 

 

 

#include"iostream"

#include"algorithm"

#include"math.h"

using namespace std;

 

//#define eps 1e-6 

 

struct Rect{

    double x1;

    double y1;

    double x2;

    double y2;

};

 

double x[300],y[300];//x和y坐标

//bool vis[300][300];

int findxi(double xx,int max)//利用线段树寻找xi和yi

{

int l=0,r=max-1,mid;

while(l<=r)

{

mid=(l+r)/2;//printf("%d %d %lf\n",l,r,xx[mid]);

if(x[mid]==xx) break;

else if(x[mid]<xx)

l=mid+1;

else

r=mid-1;

}//printf("---%d\n",mid);

return mid;

}

 

int findyi(double yy,int max)

{

int l=0,r=max-1,mid;

while(l<=r)

{

mid=(l+r)/2;//printf("%d %d %lf\n",l,r,xx[mid]);

if(y[mid]==yy) break;

else if(y[mid]<yy)

l=mid+1;

else if(y[mid]>yy)

r=mid-1;

}//printf("---%d\n",mid);

return mid;

}

 

int main(){

int Case=0;

int n=0;//多边形数目

int xy[300][300];//保存离散化所得坐标：xy[i][j]=0则空白，xy[i][j]=1则填充

Rect rect[300];//=new Rect[n];//多边形数组

int num=0;//点数目

while(scanf("%d",&n)&&n){

    //cout<<"n: "<<n<<endl;

    Case++;

    //初始化

    num=0;

    memset(rect,0,sizeof(rect));

    memset(x,0,sizeof(x));

    memset(y,0,sizeof(y));

    //读入多边形并将顶点x和y坐标分别保存在x[]和y[]数组（x[]和y[]数组后期需要排序和离散化处理）

    for(int i=0;i<n;i++){

    cin>>rect[i].x1>>rect[i].y1>>rect[i].x2>>rect[i].y2;

    //检测调整多边形左上角和右下角断点排序             

    x[num]=rect[i].x1;y[num]=rect[i].y1;

    num++;

    x[num]=rect[i].x2;y[num]=rect[i].y2;

    num++;

    }

    //cout<<"num: "<<num<<endl;//测试num是不是n的两倍

    //对x[]和y[]进行排序

    sort(x,x+num);

    sort(y,y+num);

    //对x[]和y[]做离散化处理，处理xy[]赋值

    memset(xy,0,sizeof(xy));

    int i1,i2;

    int j1,j2;

    for(int index=0;index<n;index++){//外层遍历多边形

        i1=findxi(rect[index].x1,num);

        i2=findxi(rect[index].x2,num);

        j1=findyi(rect[index].y1,num);

        j2=findyi(rect[index].y2,num);

        /*

    //内层分别遍历多边形4个点，与x[]和y[]进行index匹配并对xy[][]的值进行标注

        for(i1=0;i1<num;i1++)

            if(fabs(rect[index].x1-x[i1])<=eps)

                break;

        for(i2=0;i2<num;i2++)

            if(fabs(rect[index].x2-x[i2])<=eps)

                break;

        for(j1=0;j1<num;j1++)

            if(fabs(rect[index].y1-y[j1])<=eps)

                break;

        for(j2=0;j2<num;j2++)

            if(fabs(rect[index].y2-y[j2])<=eps)

                break;

                */

        for(int i=i1;i<i2;i++)//对xy[][]进行标注

            for(int j=j1;j<j2;j++)

                xy[i][j]=1;

 

    }//end for

    /*

    for(int i=0;i<num;i++){

        for(int j=0;j<num;j++)

            cout<<xy[i][j]<<" ";

        cout<<endl;

    }

    */

    //处理sum

    double sum=0.0;//结果

    for(int i=0;i<num;i++)

        for(int j=0;j<num;j++)

            sum+=xy[i][j]*(x[i+1]-x[i])*(y[j+1]-y[j]);

    printf("Test case #%d\n",Case);

    printf("Total explored area: %.2f\n\n",sum);

}

}