hdu 1542 （线段树求矩形面积并）

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

代码如下

#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#define lson l , m , rt << 1  
#define rson m + 1 , r , rt << 1 | 1  
using namespace std;  
const int MAXN=2222;  

double X[MAXN];  
struct node  
{  
    double l,r,h;  
    int d;  
    node(){}  
    node(double a,double b,double c,int d): l(a),r(b),h(c),d(d){}  
    bool operator <(const node &b)const  
    {  
        return h<b.h;  
    }  
}nodes[MAXN];  
int cnt[MAXN*4];  
double sum[MAXN*4];  
void PushDown(int l,int r,int rt)  
{  
    int m=(l+r)/2;  
    if(cnt[rt]!=-1)  
    {  
        cnt[rt*2]=cnt[rt*2+1]=cnt[rt];  
        sum[rt*2]= (cnt[rt]?(X[m+1]-X[l]):0) ;  
        sum[rt*2+1]= (cnt[rt]?(X[r+1]-X[m+1]):0) ;  
    }  
}  
void PushUp(int l,int r,int rt)  
{  
    if(cnt[rt*2]==-1 || cnt[rt*2+1]==-1)  
        cnt[rt]=-1;  
    else if(cnt[rt*2] != cnt[rt*2+1])  
        cnt[rt]=-1;  
    else  
        cnt[rt]=cnt[rt*2];  
    sum[rt]=sum[rt*2]+sum[rt*2+1];  
}  
void build(int l,int r,int rt)  
{  
    if(l==r)  
    {  
        cnt[rt]=0;  
        sum[rt]=0.0;  
        return ;  
    }  
    int m=(l+r)/2;  
    build(lson);  
    build(rson);  
    PushUp(l,r,rt);  
}  
void update(int L,int R,int v,int l,int r,int rt)  
{  
    if(L<=l && r<=R)  
    {  
        if(cnt[rt]!=-1)  
        {  
            cnt[rt]+=v;  
            sum[rt] = (cnt[rt]? (X[r+1]-X[l]):0);  
            return ;  
        }  
    }  
    PushDown(l,r,rt);  
    int m=(l+r)/2;  
    if(L<=m) update(L,R,v,lson);  
    if(m<R) update(L,R,v,rson);  
    PushUp(l,r,rt);  
}  
int bin(double key,int n,double d[])  
{  
    int l=1,r=n;  
    while(r>=l)  
    {  
        int m=(r+l)/2;  
        if(d[m]==key)  
            return m;  
        else if(d[m]>key)  
            r=m-1;  
        else  
            l=m+1;  
    }  
    return -1;  
}  
int main()  
{  
    int q;  
    int kase=0;  
    while(scanf("%d",&q)==1&&q)  
    {  
        int n=0,m=0;  
        for(int i=1;i<=q;i++)  
        {  
            double x1,y1,x2,y2;  
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);  
            X[++n]=x1;  
            nodes[++m]=node(x1,x2,y1,1);  
            X[++n]=x2;  
            nodes[++m]=node(x1,x2,y2,-1);  
        }  
        sort(X+1,X+n+1);  
        sort(nodes+1,nodes+m+1);  
        int k=1;//共k个不同的x坐标,组成了k-1个不同的区域  
        for(int i=2;i<=n;i++)  
            if(X[i]!=X[i-1]) X[++k]=X[i];  

        build(1,k-1,1); 
        double ret=0.0;//最终面积  
        for(int i=1;i<m;i++)  
        {  
            int l=bin(nodes[i].l,k,X);  
            int r=bin(nodes[i].r,k,X)-1;  
            if(l<=r) update(l,r,nodes[i].d,1,k-1,1);  
            ret += sum[1]*(nodes[i+1].h-nodes[i].h);  
        }  
        printf("Test case #%d\nTotal explored area: %.2lf\n\n",++kase,ret );  
    }  
}