HDU 1542 Atlantis HDU 1255 覆盖的面积扫描线/扫描线进阶

最新推荐文章于 2020-01-14 19:24:46 发布

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最新推荐文章于 2020-01-14 19:24:46 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43840511/article/details/99106760

本文介绍了一种在二维空间中计算被多个矩形区域探索总面积的算法。通过使用线段树和扫描线技术，该算法能够高效地处理大量矩形区域的重叠问题，为地图探索、地理信息系统等应用提供解决方案。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <string>
#include <vector>
//#include <iostream>
using namespace std;

struct
{
    double x0,y0,x1,y1;
}square[220];


struct inf
{
    double ll,rr,hh;
    int key;
}line[440];

int cmp(struct inf q,struct inf w)
{
    return q.hh<w.hh;
}

double x[440];
double tree[5200];
int treekey[5200];

int change(int l,int r,double ll,double rr,int now,int key)
{
    if(l+1==r)
    {
        treekey[now]+=key;
        if(treekey[now]!=0)
        {
            tree[now]=x[r]-x[l];
        }
        else
        {
            tree[now]=0;
        }
        return 0;
    }

    int mid=(l+r)/2;

    if(rr<=x[mid])
    {
        change( l, mid, ll, rr, now*2, key);
    }
    else if(ll>=x[mid])
    {
        change( mid, r, ll, rr, now*2+1, key);
    }
    else
    {
        change( l, mid, ll, rr, now*2, key);
        change( mid, r, ll, rr, now*2+1, key);
    }

    tree[now]=tree[now*2]+tree[now*2+1];

    return 0;
}

int main()
{
    int i,j,n,m,jishu=0;

    while(1)
    {
        scanf("%d",&n);

        jishu++;

        if(n==0)
        break;

        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&square[i].x0,&square[i].y0,&square[i].x1,&square[i].y1);

            x[i]=square[i].x0;
            x[i+n]=square[i].x1;

            line[i].ll=square[i].x0;
            line[i].rr=square[i].x1;
            line[i].hh=square[i].y0;
            line[i].key=1;

            line[i+n].ll=square[i].x0;
            line[i+n].rr=square[i].x1;
            line[i+n].hh=square[i].y1;
            line[i+n].key=-1;
        }

        sort(x,x+2*n);
        sort(line,line+2*n,cmp);

        m=unique(x,x+2*n)-x;



        double ans=0;
        for(i=0;i<2*n-1;i++)
        {
            change(0,m-1,line[i].ll,line[i].rr,1,line[i].key);

            ans+=(line[i+1].hh-line[i].hh)*tree[1];
        }



        printf("Test case #%d\nTotal explored area: %0.2lf\n\n",jishu,ans);



        memset(tree,0,sizeof tree);
        memset(treekey,0,sizeof treekey);
        memset(x,0,sizeof x);
        memset(line,0,sizeof line);

    }




    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
#include <string>
#include <vector>
//#include <iostream>
using namespace std;

struct
{
    double x0,y0,x1,y1;
}square[1010];


struct inf
{
    double ll,rr,hh;
    int key;
}line[2010];

int cmp(struct inf q,struct inf w)
{
    return q.hh<w.hh;
}

double x[2010];
double tree[8200];
double trees[8200];
int treekey[8200];

int change(int l,int r,double ll,double rr,int now,int key)
{
    if(x[l]>=ll&&rr>=x[r])
    {
        treekey[now]+=key;

        if(treekey[now])
        {
            tree[now]=x[r]-x[l];
        }
        else if(l==r) tree[now]=0;
        else
            tree[now]=tree[now*2]+tree[now*2+1];

        if(treekey[now]>1)
        {
            trees[now]=x[r]-x[l];
        }
        else if(l==r) trees[now]=0;
        else if(treekey[now]==1)
        {
            trees[now]=tree[now*2]+tree[now*2+1];
        }
        else trees[now]=trees[now*2]+trees[now*2+1];

        return 0;
    }

    int mid=(l+r)/2;

    if(rr<=x[mid])
    {
        change( l, mid, ll, rr, now*2, key);
    }
    else if(ll>=x[mid])
    {
        change( mid, r, ll, rr, now*2+1, key);
    }
    else
    {
        change( l, mid, ll, rr, now*2, key);
        change( mid, r, ll, rr, now*2+1, key);
    }

        if(treekey[now])
        {
            tree[now]=x[r]-x[l];
        }
        else if(l==r) tree[now]=0;
        else
            tree[now]=tree[now*2]+tree[now*2+1];

        if(treekey[now]>1)
        {
            trees[now]=x[r]-x[l];
        }
        else if(l==r) trees[now]=0;
        else if(treekey[now]==1)
        {
            trees[now]=tree[now*2]+tree[now*2+1];
        }
        else trees[now]=trees[now*2]+trees[now*2+1];

    return 0;
}

int main()
{
    int i,j,n,m,jishu=0;

    scanf("%d",&jishu);

    while(jishu--)
    {
        scanf("%d",&n);

        if(n==0)
        break;

        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&square[i].x0,&square[i].y0,&square[i].x1,&square[i].y1);

            x[i]=square[i].x0;
            x[i+n]=square[i].x1;

            line[i].ll=square[i].x0;
            line[i].rr=square[i].x1;
            line[i].hh=square[i].y0;
            line[i].key=1;

            line[i+n].ll=square[i].x0;
            line[i+n].rr=square[i].x1;
            line[i+n].hh=square[i].y1;
            line[i+n].key=-1;
        }

        sort(x,x+2*n);
        sort(line,line+2*n,cmp);

        m=unique(x,x+2*n)-x;



        double ans=0;
        for(i=0;i<2*n-1;i++)
        {
            change(0,m-1,line[i].ll,line[i].rr,1,line[i].key);

            ans+=(line[i+1].hh-line[i].hh)*trees[1];
        }



        printf("%0.2lf\n",ans);



        memset(tree,0,sizeof tree);
        memset(trees,0,sizeof tree);
        memset(treekey,0,sizeof treekey);
        memset(x,0,sizeof x);
        memset(line,0,sizeof line);

    }




    return 0;
}