hdu1542 Atlantis (线段树+矩阵面积并+离散化）

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. 

InputThe 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.OutputFor 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+矩阵面积并，被离散化难到了，之前一直没想到离散化要弄成一个左开右闭的区间

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 1e5+10;
struct seg{
    double l,r,h;
    int s;
    seg(){}
    seg(double a,double b,double c,int d):l(a),r(b),h(c),s(d){}
    bool operator < (const seg &cmp) const {
        return h < cmp.h;
    }
}t[M];
double sum[M<<2],x[M<<2];
int cnt[M<<2];
void pushup(int l,int r,int rt){
    if(cnt[rt]) sum[rt] = x[r+1] - x[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 c,int l,int r,int rt){
    if(L <= l&&R >= r){
        cnt[rt] += c;
        pushup(l,r,rt);
        return ;
    }
    mid;
    if(L <= m) update(L,R,c,lson);
    if(R > m) update(L,R,c,rson);
    pushup(l,r,rt);
}

int bin(double key,int n,double x[]){
    int l = 0;int r = n-1;
    while(l <= r){
        mid;
        if(x[m] == key) return m;
        else if(x[m] < key) l = m+1;
        else r = m-1;
    }
    return -1;
}
int main()
{
    int n,cas = 1;
    double a,b,c,d;
     while(~scanf("%d",&n)&&n){
        int m = 0;
        while(n--){
            cin>>a>>b>>c>>d;
            x[m] = a;
            t[m++] = seg(a,c,b,1);
            x[m] = c;
            t[m++] = seg(a,c,d,-1);
        }
        sort(x,x+m);
        sort(t,t+m);
        int nn = 1;
        for(int i = 1;i < m;i++){
            if(x[i]!=x[i-1]) x[nn++] = x[i];
        }
        //for(int i = 0;i < nn;i ++)
           // cout<<x[i]<<" ";
        //cout<<endl;
        double ret = 0;
        for(int i = 0;i < m-1;i ++){
            int l = bin(t[i].l,nn,x);
            int r = bin(t[i].r,nn,x)-1;
            //cout<<t[i].l<<" "<<t[i].r<<endl;
            //cout<<"l: "<<l<<" r: "<<r<<endl;
            if(l <= r) update(l,r,t[i].s,0,nn-1,1);
            //cout<<sum[1]<<endl;
            ret += sum[1] * (t[i+1].h - t[i].h);
        }
        printf("Test case #%d
Total explored area: %.2lf

",cas++ , ret);
        memset(cnt,0,sizeof(cnt));
        memset(sum,0,sizeof(sum));
     }
     return 0;
}