好久没写过博客了，这学期不是很有热情去写博客，写过的题也懒得写题解。现在来水一水博客，写一下若干年前的题目的题解。

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21978    Accepted Submission(s): 8714

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

 

 

题目就是求矩形面积并。扫描线。

一开始不知道扫描线是个什么东西的时候，感觉很厉害的东西，知道是什么之后，就。。。

就是区间更新的问题，因为在这里线段树维护的是线段，不是点，所以每个点代表的其实是一条线段，所以会有什么+1，-1的操作。

 

具体的看代码。

 

代码:

1 #include
      
        2 using namespace std; 3 typedef long long ll; 4 const int maxn=2e4+10; 5 #define lson l,m,rt<<1 6 #define rson m+1,r,rt<<1|1 7  8 int cnt[maxn<<2]; 9 double sum[maxn<<2],f[maxn];10 11 void init()12 {13     memset(cnt,0,sizeof cnt);14     memset(sum,0,sizeof sum);15 }16 17 struct node{18     double h,l,r;19     int s;20 21     node(){}22 23     node(double a,double b,double c,int d):l(a),r(b),h(c),s(d){}24 25     bool operator<(const node&cmp){26         return h
       
        >1;53     if(L<=m) update(L,R,c,lson);54     if(R> m) update(L,R,c,rson);55     pushup(rt,l,r);56 }57 58 int main()59 {60     int n,cas=1;61     while(~scanf("%d",&n)&&n){62         int h=0;63         for(int i=1;i<=n;i++){
    //从下往上扫64             double a,b,c,d;65             scanf("%lf%lf%lf%lf",&a,&b,&c,&d);66             f[h]=a;67             line[h++]=node(a,c,b,1);//标记入边68             f[h]=c;69             line[h++]=node(a,c,d,-1);//标记出边70         }71         sort(f,f+h);72         sort(line,line+h);73         int d=unique(f,f+h)-f;//离散化74         init();75         double ret=0;76         for(int i=0;i
       
      

 

 

开溜。