POJ1151(求矩阵的并面积) 离散化

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find函数直接用了线性搜索，居然没有TLE。
#include <iostream>#include <algorithm>#include <map>#include <cstdio>using namespace std;struct rec{double x1;double y1;double x2;double y2;};int find(double aim,double* arr,int T){int i;for (i=0;i<=T-1;i++){if (aim==arr[i])  return i;}}int main(){  int T;int num=0;while (1){num++;int i;int j;int k;double ans=0;int posx1;int posx2;int posy1;int posy2;int counterx=0;int countery=0;int total=0;int cover[200][200]={{0}};double x[200];double y[200];rec rectangle[100];map<double,bool> rx,ry;map<double,bool>::iterator iter;cin >> T;if (num!=1 && T!=0)cout << endl << endl;if (T==0)break;for (i=0;i<=T-1;i++){cin >> rectangle[i].x1 >> rectangle[i].y1;cin >> rectangle[i].x2 >> rectangle[i].y2;//重复的坐标不应该添加进数组//true表示已存在,flase表示不存在iter=rx.find(rectangle[i].x1);if (iter==rx.end() || iter->second==false){    x[counterx]=rectangle[i].x1;rx[rectangle[i].x1]=true;counterx++;}iter=ry.find(rectangle[i].y1);if (iter==ry.end() || iter->second==false){  y[countery]=rectangle[i].y1;ry[rectangle[i].y1]=true;countery++;}iter=rx.find(rectangle[i].x2);if (iter==rx.end() || iter->second==false){  x[counterx]=rectangle[i].x2;rx[rectangle[i].x2]=true;counterx++;}iter=ry.find(rectangle[i].y2);if (iter==ry.end() || iter->second==false){  y[countery]=rectangle[i].y2;ry[rectangle[i].y2]=true;countery++;}}sort(x,x+counterx);sort(y,y+countery);for (i=0;i<=T-1;i++){//对于每个矩阵找到坐标在数组中的位置  posx1=find(rectangle[i].x1,x,counterx);posx2=find(rectangle[i].x2,x,counterx);posy1=find(rectangle[i].y1,y,countery);posy2=find(rectangle[i].y2,y,countery);for (j=posx1;j<posx2;j++)for (k=posy1;k<posy2;k++){//第j行第k列的矩阵,1表示已覆盖cover[j][k]=1;}}for (i=0;i<=counterx-1;i++)for (j=0;j<=countery-1;j++){  if (cover[i][j]==1){ans+=(x[i+1]-x[i])*(y[j+1]-y[j]);}}cout << "Test case #" << num << endl;cout << "Total explored area: ";printf("%.2f",ans);}return 0;}