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=;

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

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

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