poj 1151 Atlantis

http://poj.org/problem?id=1151

这道题就是给你一些矩形的左上角和右下角的坐标，这些矩形可能有重叠，求这些矩形覆盖的面积。先把x坐标和y坐标分别离散化。然后再求面积。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 510
using namespace std;

int n;
double x2,y2,x1,y1;
bool flag[maxn][maxn];
double X[maxn],Y[maxn];
struct node
{
    double x1,y1,x2,y2;
} p[maxn];

int bsearch(double *a,int l,int r,double target)
{
    int low=l,high=r;
    while(low<=high)
    {
        int mid=(low+high)>>1;
        if(a[mid]==target)
        {
            return mid;
        }
        if(a[mid]>target)
            high=mid-1;
        else
            low=mid+1;
    }
}

int main()
{
    int case1=0;
    while(scanf("%d",&n)!=EOF)
    {
        memset(flag,false,sizeof(flag));
        memset(X,0,sizeof(X));
        memset(Y,0,sizeof(Y));
        case1++;
        if(n==0) break;
        int t1=0,t2=0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
            X[t1++]=p[i].x1;X[t1++]=p[i].x2;
            Y[t2++]=p[i].y1;Y[t2++]=p[i].y2;
        }
        sort(X,X+2*n);
        sort(Y,Y+2*n);
        for(int i=0; i<n; i++)
        {
            int xpos=bsearch(X,0,t1-1,p[i].x1);
            int ypos=bsearch(Y,0,t2-1,p[i].y1);
            int xpos1=bsearch(X,0,t1-1,p[i].x2);
            int ypos1=bsearch(Y,0,t2-1,p[i].y2);
            for(int i=xpos; i<xpos1; i++)
            {
                for(int j=ypos; j<ypos1; j++)
                {
                    flag[i][j]=true;
                }
            }
        }
        double sum=0;
        for(int i=0; i<t1; i++)
        {
            for(int j=0; j<t2; j++)
            {
                if(flag[i][j])
                    sum+=((X[i+1]-X[i])*(Y[j+1]-Y[j]));
            }
        }
        printf("Test case #%d
",case1);
        printf("Total explored area: %.2lf
",sum);
        printf("
");
    }
    return 0;
}