Codeforces GYM 100651 D I conduit! (Water calculation geometry)

Approximate test instructions:

1E3 segments, drawn on a piece of paper, to find out how many segments, (two segments overlap, or end-to-end will be treated as a line)

Ideas:

Two segments on the same line: they have the same slope, and they are equal to the projection points on the y-axis or x-axis. And then according to the two order of the sequence can be made out.

This card accuracy, to use the EPS

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <iostream> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include < string> #include <vector> #include <cstdio> #include <ctime> #include <bitset> #include < Algorithm> #define SZ (x) ((int) (x). Size ()) #define ALL (v) (v). Begin (), (v). End () #define A foreach (I, v) for (__typeof (v ). Begin ()) I = (v). Begin (); I! = (v). End (); + + i) #define Reveach (i, v) for (__typeof ((v). Rbegin ()) i = (v). Rbegin (); I! = (v). rend (); + + i) #define REP (i,n) for (int I=1; I<=int (n); i++) #define REP (i,n) for (int i=0; i< int (n); i++) using namespace std;typedef long long ll; #define X First#define Y    Secondtypedef pair<double,double> pii;template <class t>inline bool RD (T &ret) {char c; int sgn;    if (c = GetChar (), c = = EOF) return 0;    while (c! = '-'-' && (c< ' 0 ' | | c> ' 9 ')) C = GetChar (); Sgn= (c = = '-')?    -1:1; ret = (c = = '-')?    0: (C-' 0 ');    while (c = GetChar (), C >= ' 0 ' &&c <= ' 9 ') ret = ret * + (C-' 0 ');    RET *= SGN; return 1;}        Template <class t>inline void PT (T x) {if (x < 0) {Putchar ('-');    x =-X;    } if (x > 9) PT (X/10); Putchar (x 10 + ' 0 ');}        const int N = 1e4+100;const double inf = 1e30;const double EPS = 1e-6;int n;struct node{PII u,v; Double k,p;} All[n];bool D1 (double x,double y) {return x > y + eps;}//x > Ybool d2 (double x,double y) {return x < y- EPS;}  X < Ybool D3 (double x,double y) {return x > y-eps;}//X >= ybool D4 (double x,double y) {return x <     y + eps;}  X <= ybool dd (double x,double y) {return fabs (x-y) < EPS;}   x = = Ybool cmp (const node& A, const node& b) {if (DD (A.K, B.K)) {if (DD (A.P, B.P) {if (DD (a.u.x, b.u.x)) return D2 (A.U.Y, B.U.                        Y);                Return D2 (a.u.x, b.u.x);        } return D2 (A.P, B.P); } return D2 (A.K, B.K);} int main () {while (scanf ("%d", &n) && N) {REP (i,n) {scanf ("%lf%lf%                        Lf%lf ", &all[i].u.x, &all[i].u.y, &all[i].v.x, &AMP;ALL[I].V.Y);                        if (D1 (all[i].u.x, all[i].v.x)) swap (all[i].u, ALL[I].V); if (DD (all[i].u.x, all[i].v.x)) {if (D1 (ALL[I].U.Y, all[i].v.y)) swap (ALL[I].U, AL                                L[I].V);                                ALL[I].K = inf;                        ALL[I].P = all[i].u.x; } else {all[i].k = ((ALL[I].U.Y-ALL[I].V.Y)/(All[i].u.x-all                                [i].v.x)];                        ALL[I].P = ALL[I].U.Y-ALL[I].K * all[i].u.x; }} sort (All+1,all+1+n,CMP);                ALL[0].K = ALL[0].P =-inf;                int ans = 0, pos = 0;                Double bord;                        REP (i,n) {if (DD (ALL[I].K, INF)) {pos = i; break;}                                 if (DD (ALL[I].K, ALL[I-1].K) = = False | | DD (ALL[I].P, ALL[I-1].P) = = False) {ans + +;                        bord = all[i].v.x;                                }else{if (D1 (all[i].u.x, bord)) ans + +;                        Bord = Max (bord, all[i].v.x);                        }} if (pos) {ans + +;                        bord = all[pos].v.y; for (int i = pos+1; I <= n; i + +) {if (DD (ALL[I].P, ALL[I-1].P) = = false) ans + +, Bor                                D = all[i].v.y;                else{if (D1 (ALL[I].U.Y, bord)) ans + +;                        Bord = Max (bord, all[i].v.y);        }}}, PT (ans), puts (""); }}

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Codeforces GYM 100651 D I conduit! (Water calculation geometry)