首页 > 代码库 > 【POJ】1151 Atlantis（线段树）
【POJ】1151 Atlantis（线段树）
http://poj.org/problem?id=1151
经典矩形面积并吧.....很简单我就不说了...
有个很神的地方，我脑残没想到：
将线段变成点啊QAQ这样方便计算了啊
还有个很坑的地方，为毛每一次我精确地计算过空间可就是wa....一改大就ac...我无力了..
#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }const int N=1005;#define lc x<<1#define rc x<<1|1#define lson l, mid, lc#define rson mid+1, r, rcint tot, idn;double idy[N*2];struct T { double sum; int num; }t[N*8+10];void pushup(int x, int l, int r) { if(t[x].num>0) t[x].sum=idy[r+1]-idy[l]; else t[x].sum=t[lc].sum+t[rc].sum; }void update(int L, int R, int k, int l=1, int r=tot, int x=1) { if(L<=l && r<=R) { t[x].num+=k; pushup(x, l, r); return; } int mid=(l+r)>>1; if(L<=mid) update(L, R, k, lson); if(mid<R) update(L, R, k, rson); pushup(x, l, r);}int n, ln;struct dat { double x, y[2]; int flag; }line[N*2];bool cmp(const dat &a, const dat &b) { return a.x<b.x; }void cln() { tot=tot*4+1; rep(i, tot) t[i].sum=0, t[i].num=0; tot=0; }int main() { int test=0; while(read(n), n) { ln=idn=tot=0; for1(i, 1, n) { static double x[2], y[2]; rep(k, 2) scanf("%lf%lf", &x[k], &y[k]), idy[++idn]=y[k]; ++ln; line[ln].x=x[0]; rep(k, 2) line[ln].y[k]=y[k]; line[ln].flag=1; ++ln; line[ln].x=x[1]; rep(k, 2) line[ln].y[k]=y[k]; line[ln].flag=-1; } sort(idy+1, idy+1+idn); idn=unique(idy+1, idy+1+idn)-idy-1; tot=idn-1; sort(line+1, line+1+ln, cmp); line[0].x=line[1].x; double ans=0; for1(i, 1, ln) { ans+=t[1].sum*(line[i].x-line[i-1].x); int left=lower_bound(idy+1, idy+1+idn, line[i].y[0])-idy, rigt=lower_bound(idy+1, idy+1+idn, line[i].y[1])-idy-1; update(left, rigt, line[i].flag); } printf("Test case #%d\n", ++test); printf("Total explored area: %.2f\n", ans); puts(""); cln(); } return 0;}
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 Source Mid-Central European Regional Contest 2000 |
【POJ】1151 Atlantis（线段树）
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