zoukankan      html  css  js  c++  java
  • POJ 1151 Atlantis

    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

    2
    10 10 20 20
    15 15 25 25.5
    0

    Sample Output

    Test case #1
    Total explored area: 180.00 

    思路:
    扫描线。

    从左向右扫,记左边界为(x1,y1,y2,1) 右边界 (x2, y1, y2, -1)
    将 2*n 条线排序,2*n 个纵坐标离散化之后
    记 数组c 保存第i个区间被覆盖的次数,若被覆盖数>0,则 ans += (x[i]-x[i-1])*(y2-y1);

    用线段树保存数组c,同时优化y1,y2

    代码如下:
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <functional>
    #include <iostream>
    
    using namespace std;
    const int maxn=220;
    
    struct Line{
        double x,l,r;
        int k;
        inline bool operator<(const Line &rhs) const {
            return x<rhs.x;
        }
    }line[maxn];
    
    struct node{
        double x,l,r,len;
        bool fg;
        int cnt;
    }tr[maxn<<2];
    
    double val[maxn];
    void pushup(int i){    tr[i].len=tr[i<<1].len+tr[i<<1|1].len;}
    void build(int l,int r,int i){
        tr[i].l=val[l],tr[i].r=val[r];
        tr[i].x=0;
        tr[i].cnt=0;
        tr[i].fg=false;
        tr[i].len=tr[i].r-tr[i].l;
        if(l+1==r) {
            tr[i].fg=true;
            return;
        }
        int mid=l+r>>1;
        build(l,mid,i<<1);
        build(mid,r,i<<1|1);
        pushup(i);
    }
    double update(double x,double l,double r,int k,int i){
        if(r<=tr[i].l || l>=tr[i].r) return 0;
        if(tr[i].fg){
            if(tr[i].cnt){
                double ans=(x-tr[i].x)*(tr[i].r-tr[i].l);
                tr[i].x=x;
                tr[i].cnt+=k;
                return ans;
            } else {
                tr[i].cnt+=k;
                tr[i].x=x;
                return 0;
            }
        }
        return update(x,l,r,k,i<<1)+update(x,l,r,k,i<<1|1);
    }
    
    int main(){
        int cas=1,n;
        while(scanf("%d",&n)==1,n){
            int cnt=0;
            for (int i=1; i<=n; i++){
                double x,xx,y,yy;
                scanf("%lf%lf%lf%lf",&x,&y,&xx,&yy);
                val[++cnt]=y;
                line[cnt]=Line{x,y,yy,1};
                val[++cnt]=yy;
                line[cnt]=Line{xx,y,yy,-1};
            }
            sort(val+1,val+1+cnt);
            sort(line+1,line+1+cnt);
            build(1,cnt,1);
            double ans=0;
            for (int i=1; i<=cnt; i++)
                ans+=update(line[i].x,line[i].l,line[i].r,line[i].k,1);
            printf("Test case #%d
    Total explored area: %.2f
    
    ",cas++,ans);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    推荐一款国内首个开源全链路压测平台
    redis 你真的懂了吗?
    吊炸天的可视化安全框架,轻松搭建自己的认证授权平台!
    一条简单的更新语句,MySQL是如何加锁的?
    mysql 表删除一半数据,表空间会变小吗?
    调研字节码插桩技术,用于系统监控设计和实现
    这个开源工具把网页变成本地应用程序
    20160924-2——mysql常见问题集锦
    20160924-1——mysql存储引擎
    20160916-4:数据恢复
  • 原文地址:https://www.cnblogs.com/acerkoo/p/9845692.html
Copyright © 2011-2022 走看看