POJ1151Atlantis【离散化+扫描线+线段树】

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Atlantis

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19315 Accepted: 7335

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.

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.

Sample Input

210 10 20 2015 15 25 25.50

Sample Output

Test case #1Total explored area: 180.00 

题意：给出几个矩形求出它们组成的面积；第一道扫描线和离散化看了一下午不过总算是懂了

参考博客：http://www.cnblogs.com/fenshen371/p/3214092.html

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#define lson l,mid,rt<<1#define rson mid,r,rt<<1|1#define MAX 105using namespace std;struct Node{int flag;double l,r,h;}A[MAX<<2];int cover[MAX<<4];double num[MAX<<2],tmp[MAX<<2],len[MAX<<4];void init(){memset(len,0,sizeof(len));memset(cover,0,sizeof(cover));}bool cmp(Node a,Node b){if(a.h==b.h)return a.flag<b.flag;return a.h<b.h;}void pushup(int l,int r,int rt){if(cover[rt])len[rt]=num[r]-num[l];else if(l+1==r)len[rt]=0;else len[rt]=len[rt<<1]+len[rt<<1|1];}void update(int a,int b,int c,int l,int r,int rt){if(a==l&&b==r){cover[rt]+=c;pushup(l,r,rt);return ;}int mid=(l+r)>>1;if(mid>=b)update(a,b,c,lson);else if(mid<=a)update(a,b,c,rson);else {update(a,mid,c,lson);update(mid,b,c,rson);}pushup(l,r,rt);}int search(double now,int n){int left=1,right=n,ans;while(left<=right){int mid=(left+right)>>1;if(num[mid]>=now){ans=mid;right=mid-1;}else {left=mid+1;}}return ans;}int main(){int n,i,j,k,id,t=1;double x1,y1,x2,y2,ans;while(scanf("%d",&n),n){for(id=0,i=0;i<n;++i){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);A[++id].l=x1;A[id].r=x2;A[id].h=y1;A[id].flag=1;tmp[id]=x1;A[++id].l=x1;A[id].r=x2;A[id].h=y2;A[id].flag=-1;tmp[id]=x2;}sort(A+1,A+id+1,cmp);sort(tmp+1,tmp+id+1);for(id=0,i=1;i<=(n<<1);++i){if(i==1||tmp[i]!=tmp[i-1])num[++id]=tmp[i];}init();ans=0;A[0].h=A[1].h;for(i=1;i<=(n<<1);++i){ans+=len[1]*(A[i].h-A[i-1].h);int l=search(A[i].l,id);int r=search(A[i].r,id);update(l,r,A[i].flag,1,id,1);}printf("Test case #%d\n",t++);printf("Total explored area: %.2lf\n\n",ans);}return 0;}