首页 > ACM题库 > HDU-杭电 > HDU 3103-Shoring Up the Levees[解题报告]HOJ
2014
03-02

HDU 3103-Shoring Up the Levees[解题报告]HOJ

Shoring Up the Levees

问题描述 :

The tiny country of Waterlogged is protected by a series of levees that form a quadrilateral as shown below:

Lawrence of Arabia

The quadrilateral is defined by four vertices. The levees partition the country into four quadrants. Each quadrant is identified by a pair of vertices representing the outside edge of that quadrant. For example, Quadrant 1 shown below is defined by the points (x1, y1) and (x2, y2) .

Lawrence of Arabia

It happens very often that the country of Waterlogged becomes flooded, and the levees need to be reinforced, but their country is poor and they have limited resources. They would like to be able to reinforce those levees that encompass the largest area first, then the next largest second, then the next largest third, and the smallest area fourth.

Help Waterlogged identify which quadrants are the largest, and the length of the levees around them.

输入:

here will be several sets of input. Each set will consist of eight real numbers, on a single line. Those numbers will represent, in order:

X1 Y1 X2 Y2 X3 Y3 X4 Y4

The four points are guaranteed to form a convex quadrilateral when taken in order — that is, there will be no concavities, and no lines crossing. Every number will be in the range from -1000.0 to 1000.0 inclusive. No Quadrant will have an area or a perimeter smaller than 0.001. End of the input will be a line with eight 0.0′s.

输出:

here will be several sets of input. Each set will consist of eight real numbers, on a single line. Those numbers will represent, in order:

X1 Y1 X2 Y2 X3 Y3 X4 Y4

The four points are guaranteed to form a convex quadrilateral when taken in order — that is, there will be no concavities, and no lines crossing. Every number will be in the range from -1000.0 to 1000.0 inclusive. No Quadrant will have an area or a perimeter smaller than 0.001. End of the input will be a line with eight 0.0′s.

样例输入:

1 2 1 5 5 2 2 0
3.5 2.2 4.8 -9.6 -1.2 -4.4 -8.9 12.4
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

样例输出:

5.100 11.459 3.400 9.045 0.900 6.659 0.600 4.876
44.548 38.972 21.982 25.997 20.342 38.374 10.038 19.043

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <complex>
#include <iomanip>

using namespace std;
typedef long double ld;
typedef complex<ld> pt;
typedef vector<pt> pol;

////////////////////////////////////////////////////////////////////////////////
// General 2D geometry, Polygon cutting, Point in polygon
////////////////////////////////////////////////////////////////////////////////
const int INF = 0x3f3f3f3f; const int MINF = 0xc0c0c0c0;
const ld EPS = 1e-9; const ld PI = acos(-1.L);
ld cp(const pt &a, const pt &b) { return a.real()*b.imag() - b.real()*a.imag();}
ld dp(const pt &a, const pt &b) { return a.real()*b.real() + a.imag()*b.imag();}
inline ld sgn(const ld& x) { return abs(x) < EPS ? 0 : x/abs(x); }
inline bool cmp_lex(const pt& a, const pt& b)
{ return a.real() < b.real() || (a.real() == b.real() && a.imag() < b.imag()); }
inline bool cmp_lex_i(const pt &a, const pt &b)
{ return a.imag() < b.imag() || (a.imag() == b.imag() && a.real() < b.real()); }

// handles ALL cases, change occurences of <= to < to exclude endpoints
bool seg_x_seg(pt a1, pt a2, pt b1, pt b2){
  //if (a1==a2 || b1==b2) return false; // uncomment to exclude endpoints
  int s1 = sgn(cp(a2 - a1, b1 - a1)), s2 = sgn(cp(a2 - a1, b2 - a1));
  int s3 = sgn(cp(b2 - b1, a1 - b1)), s4 = sgn(cp(b2 - b1, a2 - b1));
  if(!s1 && !s2 && !s3) { // collinear
    if (cmp_lex(a2, a1)) swap(a1,a2); if (cmp_lex(b2, b1)) swap(b1, b2);
    return !cmp_lex(b2, a1) && !cmp_lex(a2, b1);
    //return cmp_lex(a1, b2) && cmp_lex(b1, a2);//uncomment to exclude endpoints
  } return s1*s2 <= 0 && s3*s4 <= 0; }

inline pt line_inter(const pt &a, const pt &b, const pt &c, const pt &d){
  return a + cp(c - a, d - c)/cp(b - a, d - c) * (b - a); }

// Projection of (a -> p) to vector (a -> b), SIGNED - positive in front
inline ld proj_dist(const pt &a, const pt &b, const pt &p){
  return dp(b - a, p - a) / abs(b - a); }

// SIGNED distance. Pt on the right of vector (a -> b) will be NEGATIVE.
inline ld lp_dist(const pt &a, const pt &b, const pt &p){
  return cp(b - a, p - a) / abs(b - a); }

// Line segment (a, b) to pt p distance.
inline ld lsp_dist(const pt &a, const pt &b, const pt &p){
  return dp(b - a, p - a) > 0 && dp(a - b, p - b) > 0 ?
    abs(cp(b - a, p - a) / abs(b - a)) : min(abs(a - p), abs(b - p)); }

// Closest pt on line segment (a, b) to pt p.
inline pt lsp_closest(const pt &a, const pt &b, const pt &p){
  if (dp(b - a, p - a) > 0 && dp(a - b, p - b) > 0)
    return abs(cp(b - a, p - a))<EPS ? p : line_inter(a, b, p, p+(a-b)*pt(0,1));
  return abs(a - p) < abs(b - p) ? a : b; }

// Area of a polygon (convex or concave). Always non-negative.
ld area(pol v){ ld out=0;
  for(int i = v.size()-1, j = 0; j < v.size(); i = j++) out += cp(v[i], v[j]);
  return abs(out)/2; }

ld peri(pol v) {
    ld out=0;
    for(int i=0, j=v.size(); i<j; i++) {
        out+= abs(v[i]-v[(i+1)%j]);
    }
    return out;
}

long long round3(ld a) {
    return (int) (a*1000 + 0.5);
}

bool sort_area(pol a, pol b) {
    if(round3(area(a)) == round3(area(b))) {
        return peri(a) > peri(b);
    } else {
        return area(a) > area(b);
    }
}

int main() {
    ld ax, ay, bx, by, cx, cy, dx, dy;
    while(cin >> ax>>ay>>bx>>by>>cx>>cy>>dx>>dy && 
        (abs(ax)>EPS ||
            abs(bx)>EPS || 
            abs(cx)>EPS || 
            abs(dx)>EPS || 
            abs(ay)>EPS || 
            abs(by)>EPS || 
            abs(cy)>EPS || 
            abs(dy)>EPS)) {
        pt a(ax,ay), b(bx,by), c(cx,cy), d(dx,dy), o = line_inter(a,c,b,d);
        //cerr << a << " " << b << " " << c << " " << d << " " << o << endl;
        vector<pol>p(4);
        for(int i=0; i<4; i++) {
            p[i].push_back(o);
        }
        p[0].push_back(a);
        p[0].push_back(b);
        p[1].push_back(b);
        p[1].push_back(c);
        p[2].push_back(c);
        p[2].push_back(d);
        p[3].push_back(d);
        p[3].push_back(a);
        sort(p.begin(), p.end(), sort_area);

        for(int i=0; i<4; i++) {
            cout << fixed << setprecision(3) << area(p[i]) << " " << peri(p[i]);
            if(i<3) cout << " ";
        }cout << endl;
    }
}

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