标签:style blog http color os io for ar 2014
题意:给出矩形两对角点坐标,求矩形面积并。
解法:线段树+离散化。
每加入一个矩形,将两个y值加入yy数组以待离散化,将左边界cover值置为1,右边界置为2,离散后建立的线段树其实是以y值建的树,线段树维护两个值:cover和len,cover表示该线段区间目前被覆盖的线段数目,len表示当前已覆盖的线段长度(化为离散前的真值),每次加入一条线段,将其y_low,y_high之间的区间染上line[i].cover,再以tree[1].len乘以接下来的线段的x坐标减去当前x坐标,即计算了一部分面积。
如图情况,将会计算三次面积:
代码:
#include <iostream> #include <cmath> #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #define eps 1e-8 using namespace std; #define N 307 struct node { int cov; double len; }tree[8*N]; struct Line { double y1,y2,x; int cov; }line[N]; double yy[N]; int cmp(Line ka,Line kb) { return ka.x < kb.x; } int bsearch(int l,int r,double x) { while(l <= r) { int mid = (l+r)/2; if(fabs(yy[mid]-x) < eps) return mid; if(x > yy[mid]) l = mid+1; else r = mid-1; } return l; } void build(int l,int r,int rt) { tree[rt].cov = 0; tree[rt].len = 0; if(l == r-1) return; int mid = (l+r)/2; build(l,mid,2*rt); build(mid,r,2*rt+1); } void pushup(int l,int r,int rt) { if(tree[rt].cov) tree[rt].len = yy[r]-yy[l]; else if(l+1 == r) tree[rt].len = 0; else tree[rt].len = tree[2*rt].len + tree[2*rt+1].len; } void update(int l,int r,int aa,int bb,int cover,int rt) { if(aa <= l && bb >= r) { tree[rt].cov += cover; pushup(l,r,rt); return; } if(l+1 == r) return; int mid = (l+r)/2; if(aa <= mid) update(l,mid,aa,bb,cover,2*rt); if(bb > mid) update(mid,r,aa,bb,cover,2*rt+1); pushup(l,r,rt); } int main() { int n,m,cs = 1,i,j; double x1,y1,x2,y2; while(scanf("%d",&n)!=EOF && n) { m = 1; for(i=0;i<n;i++) { scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); line[m].x = x1,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = 1,yy[m++] = y1; line[m].x = x2,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = -1,yy[m++] = y2; } m--; sort(yy+1,yy+m+1); int cnt = 2; for(i=2;i<=m;i++) { if(yy[i] != yy[i-1]) yy[cnt++] = yy[i]; } cnt--; build(1,cnt,1); sort(line+1,line+m+1,cmp); double ans = 0.0; printf("Test case #%d\n",cs++); for(i=1;i<m;i++) { int L = bsearch(1,cnt,line[i].y1); int R = bsearch(1,cnt,line[i].y2); update(1,cnt,L,R,line[i].cov,1); ans += tree[1].len*(line[i+1].x-line[i].x); } printf("Total explored area: %0.2lf\n\n",ans); } return 0; }
POJ 1151 / HDU 1542 Atlantis 线段树求矩形面积并
标签:style blog http color os io for ar 2014
原文地址:http://www.cnblogs.com/whatbeg/p/3935119.html