HDU1542 Atlantis 基于线段树的扫描线

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该题求的是矩形的面积和（重叠部分只算一次）

思路：利用二分离散化坐标（因为是浮点数），以横轴建立坐标系，矩形的上下边的长度作为区间去构造线段树，

维护cnt[]数组（表示当前边属于下边还是上边，下边为-1，上边为1）和sum[]数组（当前区间覆盖的边的长度），

一开始我觉得很难看得懂，我也是调了别人AC的代码很久才明白的。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>  #include <iostream>  #include <cstdlib>  #include <cstring>  #include <string>#include <cstdio>  #include <climits>#include <cmath> #include <vector>#include <set>#include <queue>#include <stack>#include <map>#include <sstream>#define INF 0x3f3f3f3f#define LL long long#define fora(i,a,n) for(int i=a;i<=n;i++)#define fors(i,n,a) for(int i=n;i>=a;i--)#define sci(x) scanf("%d",&x)#define scl(x) scanf("%lld",&x)#define MAXN 224#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const double eps = 1e-8;using namespace std;int n,kk=1;struct Segment{double l,r,h;int t;Segment(double _l,double _r,double _h,double _t){l=_l;r=_r;h=_h;t=_t;}Segment(){}}segment[MAXN];double X[MAXN];int cnt[MAXN<<2];double sum[MAXN<<2];bool cmp(Segment a,Segment b){return a.h-b.h<1e-6;}int erfen(double num,int MAX){int l=0,r=MAX;while(l<r){int mid=(l+r)>>1;if(X[mid]<num) l=mid+1;else if(X[mid]==num) return mid;else r=mid-1;}}void pushup(int l,int r,int rt){if(cnt[rt]) sum[rt]=X[r+1]-X[l];else if(l==r) sum[rt]=0;else sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void update(int L,int R,int v,int l,int r,int rt){if(L<=l&&r<=R){cnt[rt]+=v;pushup(l,r,rt);return;}int m=(l+r)>>1;if(L<=m) update(L,R,v,lson);if(R>m) update(L,R,v,rson);pushup(l,r,rt);}int main() {#ifdef DIDfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifdouble x1,y1,x2,y2;while(sci(n)&&n){int k=0,z=1;memset(cnt,0,sizeof(cnt));memset(sum,0,sizeof(sum));fora(i,0,n-1){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);X[k]=x1;segment[k++]=Segment(x1,x2,y1,1);X[k]=x2;segment[k++]=Segment(x1,x2,y2,-1);}fora(i,1,k-1)if(X[i]!=X[i-1]) X[z++]=X[i];sort(X,X+z);sort(segment,segment+k,cmp);double ans=0;fora(i,0,k-2){int l=erfen(segment[i].l,z-1);int r=erfen(segment[i].r,z-1)-1;update(l,r,segment[i].t,0,z-1,1);ans+=sum[1]*(segment[i+1].h-segment[i].h);}printf("Test case #%d\n",kk++);printf("Total explored area: %.2lf\n\n",ans);}return 0;}

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