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\nTotal explored area: %.2f\n\n",cas++,ans);
    }
    return 0;
}