poj-1151-Atlantis-线段树求面积并

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很裸的线段树求面积并。

坐标需要离散化一下。

#include<stdio.h>
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<map>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define maxn 11000
#define mem(a,b) (memset(a),b,sizeof(a))
#define lmin 1
#define rmax len
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define root lmin,rmax,1
#define now l,r,rt
#define int_now int l,int r,int rt
#define INF 99999999
#define LL long long
#define mod 10007
#define eps 1e-6
#define zero(x) (fabs(x)<eps?0:x)
map<double,int>mp;
double du[4010];
int len;
struct list
{
    double x1,y1;
    double x2,y2;
}node[maxn];
struct linen
{
    double y1,y2;
    int leap;
    double x;
    friend bool operator <(const linen &a,const linen &b)
    {
        if(zero(a.x-b.x)!=0) return a.x<b.x;
        else return a.leap>b.leap;
    }
}line[maxn*2];
double num[maxn*4*4*2];
int cover[maxn*4*4*2];
void push_up(int_now)
{
    if(cover[rt]==0)
    {
        num[rt]=num[rt<<1]+num[rt<<1|1];
    }
    if(cover[rt]>=1)
    {
        num[rt]=du[r+1]-du[l];
    }
  //  cout<<rt<<" "<<du[l]<<"===="<<du[r+1]<<" "<<cover[rt]<<" "<<num[rt]<<" "<<sum[rt]<<endl;
}
void push_down(int_now)
{

}
void creat(int_now)
{
    memset(cover,0,sizeof(cover));
    memset(num,0,sizeof(num));
}
void updata(int ll,int rr,int x,int_now)
{
    if(ll>r||rr<l)return;
    if(ll<=l&&rr>=r)
    {
       // cout<<l<<" "<<r<<"-"<<du[l]<<" "<<du[r+1]<<" "<<endl;
        cover[rt]+=x;
        push_up(now);
        return;
    }
    updata(ll,rr,x,lson);
    updata(ll,rr,x,rson);
    push_up(now);
}
int main()
{
    int t,n,s;
    int cas=0;
    while(~scanf("%d",&n)&&n)
    {
        cas++;
        s=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
            du[++s]=node[i].x1;
            du[++s]=node[i].y1;
            du[++s]=node[i].x2;
            du[++s]=node[i].y2;
            line[i*2-1].x=node[i].x1;
            line[i*2-1].y1=node[i].y1;
            line[i*2-1].y2=node[i].y2;
            line[i*2-1].leap=1;
            line[i*2].x=node[i].x2;
            line[i*2].y1=node[i].y1;
            line[i*2].y2=node[i].y2;
            line[i*2].leap=-1;
        }
        sort(line+1,line+n*2+1);
        sort(du+1,du+s+1);
        du[0]=-1;
        len =0;
        mp.clear();
        for(int i=1;i<=s;i++)
        {
            if(du[i]!=du[i-1])
            {
                mp[du[i]]=++len;
                du[len]=du[i];
            }
        }
        creat(root);
        double st=0;
        double are=0.0;
        for(int i=1;i<=len;i++)
        {
           // cout<<i<<"----"<<du[i]<<endl;
        }
        len--;
       // cout<<1<<" "<<len<<" "<<1<<" "<<endl;
        for(int i=1;i<=n*2;i++)
        {
            int l,r;
            l=mp[line[i].y1];
            r=mp[line[i].y2];
            if(zero(line[i].x-st)!=0)
            {
                are+=(line[i].x-st)*num[1];
                st=line[i].x;
            }
           // cout<<line[i].x<<" "<<line[i].y1<<" "<<line[i].y2<<" "<<" "<<line[i].leap<<endl;
            updata(l,r-1,line[i].leap,root);
            //cout<<l<<" "<<r<<" "<<are<<endl;
        }
        printf("Test case #%d\n",cas);
        printf("Total explored area: %.2f\n\n",are);
    }
    return 0;
}

poj-1151-Atlantis-线段树求面积并,布布扣,bubuko.com

poj-1151-Atlantis-线段树求面积并

标签：class   blog   code   com   2014   string   

原文地址：http://blog.csdn.net/rowanhaoa/article/details/29850939