Atlantis(线段树)

http://acm.hdu.edu.cn/showproblem.php?pid=1542

参考了某某大牛的代码。确实代码飘逸，给上注释，怕自己忘了；

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10210    Accepted Submission(s): 4352

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

210 10 20 2015 15 25 25.50

 

 

Sample Output

Test case #1Total explored area: 180.00 

 

 

Source

Mid-Central European Regional Contest 2000

 

/**从下到上枚举每一条边，对x建树，维护总的边在x轴的贡献**/#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <string>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long longusing namespace std;const int SIZE=2e2+10;const int maxn=500000;const int Mod=1e8;struct Line{    double h,l,r;    int f;    bool operator<(const Line &other)const{        return h<other.h;    }}l[110];double sum[SIZE<<2];int cnt[SIZE<<2];double X[220];void pushup(int l,int r,int rt){//cnt等于0时sum表示子树（不包括根节点）的覆盖的长度，else表示整棵树被覆盖了    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 build(int l,int r,int rt){    memset(cnt,0,sizeof(cnt));    memset(sum,0,sizeof(sum));}int Bin(double key,int n,double X[]){    int l=0,r=n-1;    while(l<=r){        int m=(l+r)>>1;        if(X[m]==key)return m;        if(X[m]<key)l=m+1;        else r=m-1;    }    return -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;    }    int m=(l+r)>>1;    if(L<=m)update(L,R,c,lson);    if(R>m)update(L,R,c,rson);    pushup(l,r,rt);}int main(){    int n;    double x1,x2,y1,y2;    int cas=1;    while(scanf("%d",&n)&&n){        for(int i=0;i<n;i++){            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            l[i*2]=(Line){y1,x1,x2,1};//1是下边            l[i*2+1]=(Line){y2,x1,x2,-1};//-1是下边            X[2*i]=x1;            X[2*i+1]=x2;        }        sort(X,X+2*n);        sort(l,l+2*n);        int m=1;        for(int i=1;i<2*n;i++){            if(X[i]!=X[i-1])X[m++]=X[i];//去除重复的X，得到需要离散化的最长的边        }        build(0,m-1,1);        double ans=0;        for(int i=0;i<2*n-1;i++){//最大的那一条边无需添加            int L=Bin(l[i].l,m,X);            int R=Bin(l[i].r,m,X)-1;//bin返回的是线段树边界，lr是线段树的线段            //printf("%d %d\n",L,R);            if(L<=R)update(L,R,l[i].f,0,m-1,1);            ans+=sum[1]*(l[i+1].h-l[i].h);        }        printf("Test case #%d\nTotal explored area: %.2lf\n\n",cas++,ans);    }    return 0;}

卡了好久