poj 1151(离散化+矩形面积并)
题目链接:http://poj.org/problem?id=1151
关于离散化,这篇博客讲的很好:http://www.cppblog.com/MiYu/archive/2010/10/15/129999.aspx
我线段树还是不会写这个。。
借个图:
///离散化 #include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> #include <cmath> using namespace std; const int N = 220; const double eps = 1e-10; struct Rec{ double x1,y1; double x2,y2; }rec[N]; double x[N],y[N]; int vis[N][N]; int n,k; int cmp(double a,double b){ if(a<b) return 1; return 0; } void input(){ k = 0; for(int i=0;i<n;i++){ scanf("%lf%lf%lf%lf",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2); x[k] = rec[i].x1,y[k++] = rec[i].y1; x[k] = rec[i].x2,y[k++] = rec[i].y2; } sort(x,x+k,cmp); sort(y,y+k,cmp); } int binary1(double value){ int mid,l=0,r=k-1; while(l<r){ mid = (l+r)>>1; if(fabs(x[mid]-value)<eps) return mid; if(x[mid]<value) l = mid+1; else r = mid-1; } return l; } int binary2(double value){ int mid,l=0,r=k-1; while(l<r){ mid = (l+r)>>1; if(fabs(y[mid]-value)<eps) return mid; if(y[mid]<value) l = mid+1; else r = mid-1; } return l; } double solve(){ int t1,t2,t3,t4; for(int i=0;i<n;i++){ t1 = binary1(rec[i].x1); t2 = binary1(rec[i].x2); t3 = binary2(rec[i].y1); t4 = binary2(rec[i].y2); for(int j=t1;j<t2;j++){ for(int l = t3;l<t4;l++){ vis[j][l]=1; } } } double area = 0; for(int i=0;i<k;i++){ for(int j=0;j<k;j++){ area+=vis[i][j]*(x[i+1]-x[i])*(y[j+1]-y[j]); } } return area; } int main() { int cnt=1; while(scanf("%d",&n)!=EOF&&n){ memset(vis,0,sizeof(vis)); input(); printf("Test case #%d\nTotal explored area: %.2lf\n\n",cnt++,solve()); } return 0; }