首页 > 代码库 > hdu 1542 线段树+扫描线

hdu 1542 线段树+扫描线

2024-07-31 05:15:27   214人阅读

啦啦啦~继续学算法

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1542

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7349    Accepted Submission(s): 3231


Problem 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 file 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
2
10 10 20 20
15 15 25 25.5
0
 

Sample Output
Test case #1
Total explored area: 180.00 
参考自：http://blog.csdn.net/kk303/article/details/9493265
（1）注意double~ （2）用二分解决离散化的问题
#include <iostream>
#include <string>
#include <string.h>
#include <cstdio>
#include <algorithm>
const int N=220;
const double eps=1e-6;
using namespace std;
struct node
{
  double x1,x2,y;
  int tp;
}line[N*4];
bool cmp(node a,node b)
{
  return a.y-b.y<eps; //或者写成这样  return a.y-b.y<eps  有精度问题
}

int n,times[N*4];
double sum[N*4]; //隐式构建线段树
double xx[N*4];

int find(double x)
{
  int l=1,r=n;
  while(l<=r)
  {
   int mid=(l+r)/2;
   if(xx[mid]==x)return mid;
   else if(xx[mid]<x)l=mid+1;
   else r=mid;
  }
  return l;
}

void update(int x,int t,int l,int r,int v)
{
 if(l==r)
 {
   times[x]+=t;
   if(times[x])sum[v]=xx[x+1]-xx[x];
   else sum[v]=0;
   return;
 }
 int mid=(l+r)/2;
 if(x<=mid)update(x,t,l,mid,2*v);
 else update(x,t,mid+1,r,2*v+1);
 sum[v]=sum[2*v]+sum[2*v+1];
}

int main()
{
  int test=1;
  while(scanf("%d",&n)!=EOF)
  {
    if(n==0)break;
    memset(sum,0,sizeof(sum));
    memset(times,0,sizeof(times));
    memset(xx,0,sizeof(xx));

    int num=0;
    for(int i=1;i<=n;i++)
    {
     double x1,y1,x2,y2;
     scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
     line[++num].x1=x1;
     line[num].x2=x2;
     line[num].y=y1;
     line[num].tp=1;
     xx[num]=x1;

     line[++num].x1=x1;
     line[num].x2=x2;
     line[num].y=y2;
     line[num].tp=-1;
     xx[num]=x2;
    }
    n=n*2;
    sort(xx+1,xx+1+num);
    sort(line+1,line+1+num,cmp);
    double ans=0.0;
    for(int i=1;i<=num;i++)
    {
      ans+=sum[1]*(line[i].y-line[i-1].y);
      int l=find(line[i].x1);
      int r=find(line[i].x2)-1;
      for(int j=l;j<=r;j++)update(j,line[i].tp,1,n,1);
    }
    printf("Test case #%d\n",test++);
    printf("Total explored area: %.2lf\n",ans);
    printf("\n");
  }
  return 0;
}


hdu 1542 线段树+扫描线


联系
我们