http://poj.org/problem?id=1269
我会说这种水题我手推公式+码代码用了1.5h？
还好新的一年里1A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 pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#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 printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << ' '; cout << endl
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; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const double eps=1e-6;
struct Pt { double x, y; Pt(double _x=0, double _y=0) : x(_x), y(_y) {} };
int dcmp(double a) { if(abs(a)<eps) return 0; return a<0?-1:1; }
typedef Pt Vt;
Vt operator+ (const Pt &a, const Pt &b) { return Vt(a.x+b.x, a.y+b.y); }
Vt operator- (const Pt &a, const Pt &b) { return Vt(a.x-b.x, a.y-b.y); }
Vt operator* (const Pt &a, const double &b) { return Vt(a.x*b, a.y*b); }
bool operator== (const Pt &a, const Pt &b) { return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0; }
double Cross(Vt a, Vt b) { return a.x*b.y-b.x*a.y; }
struct Line {
Pt p; Vt v;
Line() {}
Line(Pt &a, Pt &b) { p=a; v=b-a; }
};
Pt getLLP(Line &a, Line &b) {
static Pt p, q;
static Vt u, w, v;
p=a.p; q=b.p;
v=a.v; w=b.v;
u=p-q;
double t1=Cross(w, u)/Cross(v, w);
return p+v*t1;
}
// -1:xiangjiao 0:chonghe 1:pingxing
int LineAndLine(Line &p, Line &q) {
if(dcmp(Cross(p.v, q.v))!=0) return -1;
return dcmp(Cross(q.p-p.p, q.v))==0 && dcmp(Cross(q.p-p.p, p.v))==0;
}
int main() {
int n;
while(~scanf("%d", &n)) {
puts("INTERSECTING LINES OUTPUT");
Line l[2]; Pt p[4];
while(n--) {
rep(k, 4) scanf("%lf%lf", &p[k].x, &p[k].y);
l[0]=Line(p[0], p[1]);
l[1]=Line(p[2], p[3]);
int c=LineAndLine(l[0], l[1]);
if(c==-1) { Pt pt=getLLP(l[0], l[1]); printf("POINT %.2f %.2f
", pt.x, pt.y); }
else if(c==0) puts("NONE");
else puts("LINE");
}
puts("END OF OUTPUT");
}
return 0;
}
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Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000. Input The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).
Output There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".
Sample Input 5 0 0 4 4 0 4 4 0 5 0 7 6 1 0 2 3 5 0 7 6 3 -6 4 -3 2 0 2 27 1 5 18 5 0 3 4 0 1 2 2 5 Sample Output INTERSECTING LINES OUTPUT POINT 2.00 2.00 NONE LINE POINT 2.00 5.00 POINT 1.07 2.20 END OF OUTPUT Source |