HDU 1542 [POJ 1151] Atlantis （矩形面积并）

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16139    Accepted Submission(s): 6598

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 
  
  

 

Source

Mid-Central European Regional Contest 2000

 

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题意：给你n个矩形，算他们叠在一起覆盖的面积和。

分析：由于矩形的坐标是浮点类型，为处理要先离散化成整数而后进行线段树的更新。

计算矩形面积的过程，最好自己在纸上多画一画就清楚了。

ps：按照不同颜色一个个将面积加起来就得到了最后的答案。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#define lson L, mid, (rt<<1)
#define rson mid+1, R, (rt<<1|1)
using namespace std;
const int N = 1e5+10;
struct node
{
    double l,r,h;
    int temp;
    node(){}
    node(double x1,double x2,double y,int t):l(x1),r(x2),h(y),temp(t){}
    bool operator < (const node&nod)const{
        return h < nod.h;
    }
};
double sum[N<<2];
int cnt[N<<2];
vector<node> q;
vector<double>p;
int getid(double x)
{
    return lower_bound(p.begin(),p.end(),x) - p.begin();
}

void pushup(int l,int r,int rt)
{
    if(cnt[rt]) sum[rt] = p[r+1]-p[l];
    else if(l == r) sum[rt] = 0;
    else sum[rt] = sum[(rt<<1)]+sum[(rt<<1|1)];
}

void update(int L,int R,int rt,int l,int r,int temp)///l,r为查询区间
{
    if(l <= L&&R <= r)
    {
        cnt[rt] += temp;
        pushup(L,R,rt);
        return ;
    }
    int mid = (L+R)>>1;
    if(l <= mid)    update(lson,l,r,temp);
    if(r > mid)     update(rson,l,r,temp);
    pushup(L, R, rt);
}

int main()
{
    int n,cas = 0;
    while(scanf("%d",&n)==1&&n)
    {
        p.clear();
        q.clear();
        for(int i = 1;i <= n; i++)
        {
            double x1, y1, x2, y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            q.push_back(node(x1,x2,y1,1));
            q.push_back(node(x1,x2,y2,-1));
            p.push_back(x1);
            p.push_back(x2);
        }
        sort(q.begin(),q.end());
        sort(p.begin(),p.end());
        p.erase(unique(p.begin(),p.end()),p.end());
        memset(cnt,0,sizeof(cnt));
        memset(sum,0,sizeof(sum));
        double ans = 0;
        int len = q.size();
        for(int i = 0;i < len-1; i++)
        {
            int l = getid(q[i].l);
            int r = getid(q[i].r)-1;
            if(l <= r)  update(0,p.size()-1,1, l, r,q[i].temp);
            ans += sum[1]*(q[i+1].h - q[i].h);
        }
        printf("Test case #%d\n",++cas);
        printf("Total explored area: %.2f\n\n",ans);
    }
    return 0;
}