Hướng dẫn giải của Hang động
Chỉ dùng lời giải này khi không có ý tưởng, và đừng copy-paste code từ lời giải này. Hãy tôn trọng người ra đề và người viết lời giải.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Nộp một lời giải chính thức trước khi tự giải là một hành động có thể bị ban.
Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này
Code mẫu của ladpro98
#include <cstring> #include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <climits> #include <cstdlib> #include <ctime> #include <memory.h> #include <cassert> #include <climits> #define FOR(i, a, b) for(int i = (a); i < (b); i++) #define REP(i, a, b) for(int i = (a); i <=(b); i++) #define FORD(i, a, b) for(int i = (a); i > (b); i--) #define REPD(i, a, b) for(int i = (a); i >=(b); i--) #define TR(it, a) for(typeof((a).begin()) it = (a).begin(); it != (a).end(); it++) #define RESET(a, v) memset((a), (v), sizeof((a))) #define SZ(a) (int((a).size())) #define ALL(a) (a).begin(), (a).end() #define PB push_back #define MP make_pair #define LL long long #define LD long double #define II pair<int, int> #define X first #define Y second #define VI vector<int> const int N = 5050; const double EPS = 1e-9; using namespace std; typedef pair<double, double> point; typedef pair<double, double> line; bool eq(const double &a, const double &b) {return fabs(a - b) < EPS;} bool eqX(const point &a, const point &b) {return eq(a.X, b.X);} line makeLine(point P, point Q) { line tmp; tmp.X = (P.Y - Q.Y) / (P.X - Q.X); tmp.Y = P.Y - tmp.X * P.X; return tmp; } double xIntercept(line d1, line d2) {return (d2.Y - d1.Y) / (d1.X - d2.X);} double yIntercept(line d1, line d2){ double x = xIntercept(d1, d2); return d1.X * x + d1.Y; } bool bad(line a, line b, line c) {return xIntercept(a, b) > EPS + xIntercept(b, c);} bool cmp(const point &a, const point &b) { return !eq(a.X, b.X) ? a.X < b.X : a.Y > b.Y; } int n; point a[N]; int main() { ios :: sync_with_stdio(0); cin.tie(0); cin >> n; FOR(i, 0, n) cin >> a[i].X >> a[i].Y; vector<line> lines, hull; FOR(i, 1, n) lines.PB(makeLine(a[i - 1], a[i])); sort(ALL(lines), cmp); lines.resize(unique(ALL(lines), eqX) - lines.begin()); #define nn SZ(hull) FOR(i, 0, SZ(lines)) { while (nn > 1 && bad(hull[nn - 2], hull[nn - 1], lines[i])) hull.pop_back(); hull.PB(lines[i]); } double ans = 1e7; FOR(i, 1, nn) ans = min(ans, yIntercept(hull[i - 1], hull[i])); cout << setprecision(2) << fixed << ans; return 0; }
Code mẫu của RR
{$R+,Q+,N+} PROGRAM NKSPILJA; CONST FINP=''; FOUT=''; oo=1000000; maxn=5000; max=1000000; esp=0.005; VAR x,y:array[1..maxn] of double; n:longint; kq:double; procedure readInput; var f:text; i:integer; begin assign(f,FINP); reset(f); read(f,n); for i:=1 to n do read(f,x[i],y[i]); close(f); end; procedure writeOutput; var f:text; begin assign(f,FOUT); rewrite(f); writeln(f,kq:0:2); close(f); end; function ok(y0:double):boolean; var l,r,x0:double; i:longint; begin l:=0; r:=max; for i:=1 to n-1 do if y[i]=y[i+1] then begin if y0<y[i] then begin l:=oo; r:=0; end; end else if y[i]>y[i+1] then begin x0:=(x[i]*y[i+1]-y[i]*x[i+1]-(x[i]-x[i+1])*y0)/(y[i+1]-y[i]); if x0>l then l:=x0; end else if y[i]<y[i+1] then begin x0:=(x[i]*y[i+1]-y[i]*x[i+1]-(x[i]-x[i+1])*y0)/(y[i+1]-y[i]); if x0<r then r:=x0; end; if l<r then ok:=true else ok:=false; end; procedure chia(y1,y2:double); var mid:double; begin mid:=(y1+y2)/2; if ok(mid) then begin if mid-y1<esp then begin kq:=mid; writeOutput; halt; end; chia(y1,mid); end else begin if y2-mid<esp then begin kq:=mid; writeOutput; halt; end; chia(mid,y2); end; end; BEGIN readInput; chia(0,oo); END.
Code mẫu của hieult
#include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <cassert> #include <cstdlib> #include <cstring> #include <sstream> #include <iomanip> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; typedef long double ld; //typedef double ld; typedef unsigned int ui; typedef unsigned long long ull; #define Rep(i,n) for(__typeof(n) i = 0; i < (n); ++i) #define Repd(i,n) for(__typeof(n) i = (n)-1; i >= 0; --i) #define For(i,a,b) for(__typeof(b) i = (a); i <= (b); ++i) #define Ford(i,a,b) for(__typeof(a) i = (a); i >= (b); --i) #define Fit(i,v) for(__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) #define Fitd(i,v) for(__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++i) #define mp make_pair #define pb push_back #define fi first #define se second #define sz(a) ((int)(a).size()) #define all(a) (a).begin(), (a).end() #define ms(a,x) memset(a, x, sizeof(a)) #define nl puts("") #define sp printf(" ") #define ok puts("ok") //#include <conio.h> template<class F, class T> T convert(F a, int p = -1) { stringstream ss; if (p >= 0) ss << fixed << setprecision(p); ss << a; T r; ss >> r; return r; } template<class T> void db(T a, int p = -1) { if (p >= 0) cout << fixed << setprecision(p); cout << a << " "; } template<class T> T gcd(T a, T b) { T r; while (b != 0) { r = a % b; a = b; b = r; } return a; } template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<class T> T sqr(T x) { return x * x; } template<class T> T cube(T x) { return x * x * x; } template<class T> struct Triple { T x, y, z; Triple() {} Triple(T _x, T _y, T _z) : x(_x), y(_y), z(_z) {} }; template<class T> Triple<T> euclid(T a, T b) { if (b == 0) return Triple<T>(1, 0, a); Triple<T> r = euclid(b, a % b); return Triple<T>(r.y, r.x - a / b * r.y, r.z); } template<class T> int getbit(T s, int i) { return (s >> i) & 1; } template<class T> T onbit(T s, int i) { return s | (T(1) << i); } template<class T> T offbit(T s, int i) { return s & (~(T(1) << i)); } template<class T> int cntbit(T s) { return s == 0 ? 0 : cntbit(s >> 1) + (s & 1); } const int bfsz = 1 << 16; char bf[bfsz + 5]; int rsz = 0;int ptr = 0; char gc() { if (rsz <= 0) { ptr = 0; rsz = fread(bf, 1, bfsz, stdin); if (rsz <= 0) return EOF; } --rsz; return bf[ptr++]; } void ga(char &c) { c = EOF; while (!isalpha(c)) c = gc(); } int gs(char s[]) { int l = 0; char c = gc(); while (isspace(c)) c = gc(); while (c != EOF && !isspace(c)) { s[l++] = c; c = gc(); } s[l] = '\0'; return l; } template<class T> bool gi(T &v) { v = 0; char c = gc(); while (c != EOF && c != '-' && !isdigit(c)) c = gc(); if (c == EOF) return false; bool neg = c == '-'; if (neg) c = gc(); while (isdigit(c)) { v = v * 10 + c - '0'; c = gc(); } if (neg) v = -v; return true; } const double PI = 2 * acos(0); const string months[] = {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"}; const int days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int dr[] = {-1, +0, +1, +0}; const int dc[] = {+0, +1, +0, -1}; const int inf = (int)1e9 + 5; const ll linf = (ll)1e16 + 5; const ld eps = ld(1e-9); const ll mod = 100000007; typedef pair<int, int> II; #define maxn 5005 struct Point{ ld x, y; Point(){}; Point(ld _x, ld _y){ x = _x; y = _y; } }; bool dcmp(ld u, ld v){ return abs(u - v) < eps; } double giao(Point P0, Point P1, double y){ return (P1.x - P0.x) * (y - P0.y) / (P1.y - P0.y) + P0.x; } int n; Point A[maxn]; bool thoaman(double y){ double xl = -double(linf), xr = double(linf), x; For(i, 1, n - 1){ if(!dcmp(A[i].y, A[i + 1].y)) { x = giao(A[i], A[i + 1], y); if(A[i].y > A[i + 1].y) xl = max(xl, x); else xr = min(xr, x); } else if(y + eps < A[i].y){ return false; } } return xr > xl; } int main(){ // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); cin >> n; For(i, 1, n) cin >> A[i].x >> A[i].y; ld u = 0, v = 1000001, mid; Rep(run, 100){ mid = (u + v) / 2; if(thoaman(mid)) v = mid; else u = mid; } cout << fixed << setprecision(2); cout << u; }
Code mẫu của ll931110
{$N+} program spil; const input = ''; output = ''; maxn = 5000; maxv = 1000000; maxt = 1000000000; eps = 1e-6; type point = record x,y: extended; end; line = record a,b,c: extended; end; var p: array[1..maxn] of point; d: array[1..maxn] of line; n: longint; res: extended; procedure init; var f: text; i: longint; begin assign(f, input); reset(f); readln(f, n); for i := 1 to n do readln(f, p[i].x, p[i].y); close(f); end; procedure expand(i: longint); begin with d[i] do begin a := p[i].y - p[i + 1].y; b := p[i + 1].x - p[i].x; c := -(a * p[i].x + b * p[i].y); end; end; function ok(q: extended): boolean; var low,high,tmp: extended; i: longint; begin low := -maxt; high := maxt; for i := 1 to n - 1 do if p[i].y = p[i + 1].y then begin if q < p[i].y then exit(false); if q = p[i].y then begin if low < p[i].x then low := p[i].x; if high > p[i].x then high := p[i].x; end; end else begin with d[i] do tmp := -(c + b * q)/a; if (p[i].y > p[i + 1].y) and (low < tmp) then low := tmp; if (p[i].y < p[i + 1].y) and (high > tmp) then high := tmp; end; ok := (high >= low); end; procedure solve; var inf,sup,med: extended; i: longint; begin for i := 1 to n - 1 do expand(i); inf := 0; sup := maxv; repeat med := (inf + sup) / 2; if ok(med) then begin res := med; sup := med; end else inf := med; until sup - inf < eps; end; procedure printresult; var f: text; begin assign(f, output); rewrite(f); writeln(f, res:0:2); close(f); end; begin init; solve; printresult; end.
Code mẫu của khuc_tuan
#include <iostream> using namespace std; struct Point { int x, y; Point() {} Point(int x,int y) : x(x), y(y) {} }; int n; Point a[5050]; bool checkok(double Y) { double lower = a[0].x, upper = a[n-1].x; for(int i=0;i+1<n;++i) { // a[i] -> a[i+1] if(a[i].y==a[i+1].y) { if(a[i].y > Y) return false; } else { double X = (Y - a[i+1].y) * (a[i+1].x - a[i].x) / (a[i+1].y - a[i].y) + a[i+1].x; if(a[i+1].y>a[i].y) upper = min( upper, X); else lower = max( lower, X); } } return lower <= upper; } int main() { scanf("%d", &n); for(int i=0;i<n;++i) scanf("%d%d", &a[i].x, &a[i].y); double left = 0, right = 1000000; for(int kk=0;kk<50;++kk) { double mid = (left+right) / 2; if(checkok(mid)) right = mid; else left = mid; } printf("%.2f\n", left); return 0; }
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