2017 United Kingdom and Ireland Programming Contest B 几何、点到直线距离 F 概率dp

2017 United Kingdom and Ireland Programming Contest (UKIEPC 2017)  

B    Breaking Biscuits

题意：一个多边形的饼干，要把它放进一个杯子里，可以随意旋转饼干，求杯子的最小直径。

tags：枚举多边形的所有边（即任意两个点形成的所有边），每次对这条边求两边的点到它的最远距离。放的时候就是让这条边竖直放。

点到直线距离模板

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
}a[N];
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.x*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.x/p); }
bool operator < (const Point& a, const Point b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double EPS = 1e-10;
int dcmp(double x) {
    if(fabs(x) < EPS) return 0;
    else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y);
}
//向量点积
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
//向量长度
double Length(Vector A) { return sqrt(Dot(A, A)); }
//向量叉积
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
//点到直线距离
double DistanceToLine(Point P, Point A, Point B) {
    Vector v1 = B - A, v2 = P - A;
    return Cross(v1, v2) / Length(v1); //不取绝对值，得到的是有向距离
}

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
}a[N];
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.x*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.x/p); }
bool operator < (const Point& a, const Point b) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double EPS = 1e-10;
int dcmp(double x) {
    if(fabs(x) < EPS) return 0;
    else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b) {
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y);
}
//向量点积
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
//向量长度
double Length(Vector A) { return sqrt(Dot(A, A)); }
//向量叉积
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
//点到直线距离
double DistanceToLine(Point P, Point A, Point B) {
    Vector v1 = B - A, v2 = P - A;
    return Cross(v1, v2) / Length(v1); //不取绝对值，得到的是有向距离
}

Vector  p[110];
double ans = 1e18;
int n;
int main()
{
    scanf("%d", &n);
    rep(i,1,n)
        scanf("%lf%lf", &p[i].x, &p[i].y);
    double tmp, m1=0, m2=0;
    rep(i,1,n)
    {
        rep(j,1,n) if(j!=i)
        {
            m1 = m2 =0;
            rep(k,1,n) if(k!=i && k!=j)
            {
                tmp = DistanceToLine(p[k], p[i], p[j]);
                if(tmp>0) m1=max(m1, tmp);
                else m2=min(m2, tmp);
            }
            ans = min(ans, m1-m2);
        }
    }
    printf("%.6f
", ans);

    return 0;
}

F  Flipping Coins

题意：n 枚硬币，一开始都朝下。k 次操作，每次取一枚抛，且必须抛 k 次。问在明智的前提下，最后正面朝上的硬币期望数量是多少。

tags：设 dp[i][j] 表示抛完第 i 次后，有 j 枚硬币朝上的概率。然后递推一下就是了。

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 505;

double dp[N][N];
int n, k;
int main()
{
    scanf("%d%d", &n, &k);
    dp[0][0] = 1;
    rep(i,1,k) rep(j,0,i)
    {
        if(j<n) {
            dp[i][j] += dp[i-1][j]/2;
            dp[i][j+1] += dp[i-1][j]/2;
        }
        else if(j==n) {
            dp[i][j-1] += dp[i-1][j]/2;
            dp[i][j] += dp[i-1][j]/2;
        }
    }
    double ans=0;
    rep(i,1,n) ans += dp[k][i]*i;
    printf("%.6f
", ans);

    return 0;
}