zoukankan      html  css  js  c++  java
  • hdu-1542 Atlantis(离散化+线段树+扫描线算法)

    题目链接:

    Atlantis

    Time Limit: 2000/1000 MS (Java/Others)   

     Memory Limit: 65536/32768 K (Java/Others)


    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
     
    题意:
     
    给了n个矩形的左下角和右上角的坐标,问这些矩形的面积和是多少;
     
    思路:
     
    离散化后用线段树结合扫描线算法可以做,是扫描线算法的模板题;
     
    AC代码:
     
    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int N=1e5+10;
    int n;
    double rec[2*N],xa,xb,ya,yb;
    struct Tree
    {
        int l,r,cover;
        double sum;
    };
    Tree tree[4*N];
    struct Line
    {
        double l,r,h;
        int flag;
    };
    Line line[4*N];
    int cmp(Line x,Line y)
    {
        return x.h<y.h;
    }
    void build(int node,int L,int R)
    {
        tree[node].sum=0;
        tree[node].cover=0;
        tree[node].l=L;
        tree[node].r=R;
        if(L>=R)return ;
        int mid=(L+R)>>1;
        build(2*node,L,mid);
        build(2*node+1,mid+1,R);
    }
    void Pushup(int node)
    {
        if(tree[node].cover)
        {
            tree[node].sum=rec[tree[node].r+1]-rec[tree[node].l];
        }
        else
        {
            if(tree[node].l!=tree[node].r)
            tree[node].sum=tree[2*node].sum+tree[2*node+1].sum;
            else tree[node].sum=0;
        }
    }
    void update(int node,int L,int R,int x)
    {
        if(L<=tree[node].l&&R>=tree[node].r)
        {
            tree[node].cover+=x;
            Pushup(node);
            return ;
        }
        int mid=(tree[node].l+tree[node].r)>>1;
        if(R<=mid)update(2*node,L,R,x);
        else if(L>mid)update(2*node+1,L,R,x);
        else
        {
            update(2*node,L,mid,x);
            update(2*node+1,mid+1,R,x);
        }
        Pushup(node);
    }
    map<double,int>mp;
    int main()
    {
        int Case=1;
        while(1)
        {
            scanf("%d",&n);
            if(n==0)break;
            printf("Test case #%d
    ",Case++);
            int cnt=1;
            for(int i=0;i<n;i++)
            {
                scanf("%lf%lf%lf%lf",&xa,&ya,&xb,&yb);
    
                rec[cnt]=line[cnt].l=xa;
                line[cnt].r=xb;
                line[cnt].flag=1;
                line[cnt++].h=ya;
    
                line[cnt].l=xa;
                rec[cnt]=line[cnt].r=xb;
                line[cnt].h=yb;
                line[cnt++].flag=-1;
            }
            sort(line+1,line+cnt,cmp);
            sort(rec+1,rec+cnt);
            int num=2;
            for(int i=2;i<cnt;i++)
            {
                if(rec[i]!=rec[i-1])rec[num++]=rec[i];
            }
            for(int i=1;i<num;i++)
            {
                mp[rec[i]]=i;
            }
            build(1,1,num-1);
            double ans=0;
    
            for(int i=1;i<cnt-1;i++)
            {
                int fx=mp[line[i].l];
                int fy=mp[line[i].r];
                update(1,fx,fy-1,line[i].flag);
                ans+=tree[1].sum*(line[i+1].h-line[i].h);
            }
            printf("Total explored area: %.2lf
    ",ans);
            printf("
    ");
        }
        return 0;
    }
     
  • 相关阅读:
    help python(查看模块帮助文档)
    Vim常用快捷键
    tar 解压缩
    目前的学习计划
    学习方向
    C#转Python计划
    困惑的屌丝,求方向。。。
    修改PYTHONPATH的一种方法(在Window平台和Ubuntu下都有效)
    使用正则表达式统计vs项目代码总行数[转]
    日常工作细节汇总
  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5366207.html
Copyright © 2011-2022 走看看