首页 > 代码库 > POJ1151Atlantis 矩形面积并[线段树 离散化 扫描线]
POJ1151Atlantis 矩形面积并[线段树 离散化 扫描线]
Atlantis
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 21734 | Accepted: 8179 |
Description
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
Input
The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don‘t process it.
The input file is terminated by a line containing a single 0. Don‘t process it.
Output
For each test case, your program should output one section. The first line of each section must be "Test case #k", where k is the number of the test case (starting with 1). The second one must be "Total explored area: a", where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Output a blank line after each test case.
Sample Input
210 10 20 2015 15 25 25.50
Sample Output
Test case #1Total explored area: 180.00
Source
Mid-Central European Regional Contest 2000
[10-13-2016]
有一条线从下往上扫，维护当前的x轴覆盖总长度
传送门
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;const int N=205;#define m (l+r)/2#define lson o<<1,l,m#define rson o<<1|1,m+1,r#define lc o<<1#define rc o<<1|1inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} return x*f;}int n,cnt=0;double x1,y1,x2,y2,mp[N];struct seg{ double l,r,h; int f;//1 or -1 seg(double a=0,double b=0,double c=0,int d=0):l(a),r(b),h(c),f(d){} bool operator <(const seg &r)const{return h<r.h;}}a[N];struct node{ double sum; int cov;}t[N<<2];inline int Bin(double v){ int l=1,r=cnt; while(l<=r){ int mid=(l+r)>>1; if(mp[mid]==v) return mid; else if(v<mp[mid]) r=mid-1; else l=mid+1; } return -1;}inline void pushUp(int o,int l,int r){ if(t[o].cov) t[o].sum=mp[r+1]-mp[l]; else if(l==r) t[o].sum=0; else t[o].sum=t[lc].sum+t[rc].sum;}void update(int o,int l,int r,int ql,int qr,int v){ if(ql<=l&&r<=qr){ t[o].cov+=v; pushUp(o,l,r); }else{ if(ql<=m) update(lson,ql,qr,v); if(m<qr) update(rson,ql,qr,v); pushUp(o,l,r); }}int cas=0;int main(int argc, const char * argv[]) { while((n=read())){ double ans=0; for(int i=1;i<=n;i++){ scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); a[i*2-1]=seg(x1,x2,y1,1); a[i*2]=seg(x1,x2,y2,-1); mp[2*i-1]=x1; mp[2*i]=x2; } sort(mp+1,mp+1+2*n); sort(a+1,a+1+2*n); cnt=0;mp[++cnt]=mp[1]; for(int i=2;i<=2*n;i++) if(mp[i]!=mp[i-1]) mp[++cnt]=mp[i]; memset(t,0,sizeof(t)); for(int i=1;i<=2*n-1;i++){//最后一个不用 int ql=Bin(a[i].l),qr=Bin(a[i].r)-1; if(ql<=qr) update(1,1,cnt,ql,qr,a[i].f); ans+=t[1].sum*(a[i+1].h-a[i].h); } printf("Test case #%d\n",++cas); printf("Total explored area: %.2f\n\n",ans); } return 0;}
线段树需要插入线段，删除线段，求线段覆盖的总长度，貌似还是用标记永久化比较方便，否则删(我)除(没)很(写)麻(出)烦(来)
注意这个线段树节点是一段区间哦
离散化m忘清0了 WA了几次
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;const int N=205;#define lson x<<1,l,mid#define rson x<<1|1,mid+1,r#define lc x<<1#define rc x<<1|1inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();} return x*f;}int n;double x1,y1,x2,y2;struct Seg{ double l,r,y; int f; Seg(double l=0,double r=0,double y=0,int f=0):l(l),r(r),y(y),f(f){} bool operator <(const Seg &a)const{return y<a.y;}}a[N];double mp[N];int m;void iniMP(){ sort(mp+1,mp+1+m); int p=0; mp[++p]=mp[1]; for(int i=2;i<=m;i++) if(mp[i]!=mp[i-1]) mp[++p]=mp[i]; m=p;}inline int Bin(double v){ int l=1,r=m; while(l<=r){ int mid=(l+r)>>1; if(v==mp[mid]) return mid; else if(v<mp[mid]) r=mid-1; else l=mid+1; } return 0;}struct node{ double sum; int cov; node():sum(0),cov(0){}}t[N<<2];void pushUp(int x,int l,int r){ if(t[x].cov) t[x].sum=mp[r+1]-mp[l]; else if(l==r) t[x].sum=0; else t[x].sum=t[lc].sum+t[rc].sum;}void segCov(int x,int l,int r,int ql,int qr,int v){ if(ql<=l&&r<=qr) t[x].cov+=v,pushUp(x,l,r); else{ int mid=(l+r)>>1; if(ql<=mid) segCov(lson,ql,qr,v); if(mid<qr) segCov(rson,ql,qr,v); pushUp(x,l,r); }}int cas=0;int main(int argc, const char * argv[]) { while((n=read())){ m=0; for(int i=1;i<=n;i++){ scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); a[i*2-1]=Seg(x1,x2,y1,1); a[i*2]=Seg(x1,x2,y2,-1); mp[++m]=x1;mp[++m]=x2; } iniMP(); n<<=1; sort(a+1,a+1+n); memset(t,0,sizeof(t)); double ans=0; for(int i=1;i<=n-1;i++){ int ql=Bin(a[i].l),qr=Bin(a[i].r)-1; if(ql<=qr) segCov(1,1,m,ql,qr,a[i].f); ans+=t[1].sum*(a[i+1].y-a[i].y); } printf("Test case #%d\n",++cas); printf("Total explored area: %.2f\n\n",ans); } return 0;}
POJ1151Atlantis 矩形面积并[线段树 离散化 扫描线]
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