Atlantis(坐标离散化)

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

题意：给出矩形的左上角坐标和右下角坐标（坐标的y轴是向下的），求出矩形面积的并。。

今天好困啊。。迷迷糊糊的听会神给讲了讲，敲完之后调试了好久。。原来存错数组了。。看来意识模糊的时候不适宜敲题。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=550;

struct node
{
    double x1,y1,x2,y2;
} p[N];
double X[N],Y[N];
bool vis[N][N];

void init()
{
    memset(vis,0,sizeof(vis));
    memset(X,0,sizeof(X));
    memset(Y,0,sizeof(Y));
}
int BinSearch(double *f,int l,int r,double key)
{
    while(l <= r)
    {
        int mid = (l+r)>>1;
        if (f[mid]==key)
            return mid;
        else if (key < f[mid])
            r = mid-1;
        else
            l = mid+1;
    }
}
int main()
{
    int o = 0,n;
    while(~scanf("%d",&n)&&n)
    {
        o++;
        init();
        int m1=0,m2=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[m1++] = p[i].x1;
            Y[m2++] = p[i].y1;
            X[m1++] = p[i].x2;
            Y[m2++] = p[i].y2;
        }
        sort(X,X+m1);
        sort(Y,Y+m2);
        for (int i = 0; i < n; i++)
        {
            int px1 = BinSearch(X,0,m1-1,p[i].x1);
            int py1 = BinSearch(Y,0,m2-1,p[i].y1);
            int px2 = BinSearch(X,0,m1-1,p[i].x2);
            int py2 = BinSearch(Y,0,m2-1,p[i].y2);
            for (int i = px1; i < px2; i++)
            {
                for (int j = py1; j < py2; j++)
                    vis[i][j] = true;
            }

        }
        double area = 0;
        for (int i = 0; i < m1; i++)
        {
            for (int j = 0; j < m2; j++)
            {
                if (vis[i][j])
                {
                    area += (X[i+1]-X[i])*(Y[j+1]-Y[j]);
                }
            }
        }
        printf("Test case #%d
",o);
        printf("Total explored area: %.2f
",area);
        puts("");
    }
    return 0;
}