zoj 1128 || poj 1151 Atlantis

求矩形并的面积。

 

传说中的离散化。

 

因为矩形最多才100个，所以将所有的x坐标从小到大映射下，y也是。数组map[I][K]相当于横坐标为x[I] 纵坐标为y[K]的点，也就相当于把很大距离的点缩到了200*200上，然后再扫描下求面积即可。

 

引用下PPT的这个图。。

 

 

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 205;const double eps = 1e-6;struct rectangle{ double lx,ly,rx,ry;};rectangle r[MAX];double x[MAX],y[MAX];int map[MAX][MAX];bool dy(double x,double y){return x > y + eps;}// x > y bool xy(double x,double y){return x < y - eps;}// x < y bool dyd(double x,double y){ return x > y - eps;}// x >= y bool xyd(double x,double y){return x < y + eps;} // x <= y bool dd(double x,double y) {return fabs( x - y ) < eps;} // x == yint find(double a[],double x,int m){int beg = 0,end = m-1;while( beg <= end ){int mid = (beg+end) >> 1;if( dd(x,a[mid]) )return mid;if( dy(x,a[mid]) )beg = mid + 1;elseend = mid;}}double area_rectangle(int m,int n){for(int i=0; i<n; i++){int x1 = find(x,r[i].lx,m);int y1 = find(y,r[i].ly,m);int x2 = find(x,r[i].rx,m);int y2 = find(y,r[i].ry,m);for(int k=x1; k<x2; k++)for(int j=y1; j<y2; j++)map[k][j] = true;}double sum = 0.0;for(int i=0; i<m-1; i++)for(int k=0; k<m-1; k++)if( map[i][k] )sum += (x[i+1] - x[i])*(y[k+1] - y[k]);return sum;}int main(){int n;int ind = 1;while( ~scanf("%d",&n) && n ){int k = 0;memset(map,false,sizeof(map));for(int i=0; i<n; i++){scanf("%lf%lf%lf%lf",&r[i].lx,&r[i].ly,&r[i].rx,&r[i].ry);x[k] = r[i].lx;y[k++] = r[i].ly;x[k] = r[i].rx;y[k++] = r[i].ry;}sort(x,x+k);sort(y,y+k);double area = area_rectangle(k,n);printf("Test case #%d/nTotal explored area: %.2lf/n/n",ind++,area);}return 0;}