POJ 1151 Atlantis

Atlantis

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1151

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 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 

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<iostream>#include<algorithm>using namespace std;#define N 300struct node{    int l,r,c;    double lf,rf,cnt;}q[N<<2];struct tt{    double x,y1,y2;    int v;}p[N<<2];double pp[N<<2];bool cmp(tt a, tt b){    return a.x < b.x;}void build(int l,int r,int rt){    q[rt].l = l;    q[rt].r = r;    q[rt].c = 0;    q[rt].cnt = 0;    q[rt].lf =  pp[l];    q[rt].rf = pp[r];    if(l+1 == r)    {        return ;    }    int mid = (l+r)>>1;    build(l,mid,rt<<1);    build(mid,r,rt<<1|1);}void geng(int rt){    if(q[rt].c>0)    {        q[rt].cnt = q[rt].rf - q[rt].lf;        return ;    }    if(q[rt].l+1 == q[rt].r)    {        q[rt].cnt = 0;    }    else    {        q[rt].cnt = q[rt<<1].cnt + q[rt<<1|1].cnt;    }}void insert(int rt,tt e){    if(e.y1 == q[rt].lf && e.y2 == q[rt].rf)    {        q[rt].c = q[rt].c + e.v;        geng(rt);        return ;    }    if(e.y2<=q[rt<<1].rf)    {        insert(rt<<1,e);    }    else if(e.y1>=q[rt<<1|1].lf)    {        insert(rt<<1|1,e);    }    else    {        tt w = e;        w.y2 = q[rt<<1].rf;        insert(rt<<1,w);        w = e;        w.y1 = q[rt<<1|1].lf;        insert(rt<<1|1,w);    }    geng(rt);}int main(){    int n,m,i,j;    int kk = 0;    while(scanf("%d",&n)!=EOF)    {        kk++;        if(n == 0)        {            break;        }        double a,b,c,d;        int m = 1;        for(i=1;i<=n;i++)        {            scanf("%lf%lf%lf%lf",&a,&b,&c,&d);            p[m].x = a;            p[m].y1 = b;            p[m].y2 = d;            p[m].v = 1;            pp[m++] = b;            p[m].x = c;            p[m].y1 = b;            p[m].y2 = d;            p[m].v = -1;            pp[m++] = d;        }        m--;        sort(p+1, p+m+1, cmp);          sort(pp+1, pp+m+1);         build(1,m,1);        insert(1,p[1]);        double count = 0;        for(i=2;i<=m;i++)        {            count = count + q[1].cnt *(p[i].x-p[i-1].x);            insert(1,p[i]);        }        printf("Test case #%d\n",kk);        printf("Total explored area: %.2lf\n\n",count);    }    return 0;}