hdu 5473 There was a kingdom（dp+幾何）

題目連結：hdu 5473 There was a kingdom

解題思路

選取的點一定在凸包上，是以對點集做凸包，如果凸包的點個數小于等于K，面積可以取到最大值。否則，枚舉起點，做動态規劃。dp[i][j]表示到第i個點選取了j個點的最優解。這樣的複雜度為 o(n2k) ，算上枚舉起點總得複雜度為 o(n3k) 。但是，我們選取的是k個點，如果有枚舉到一次的起點在這k個點上，即可以命中最優答案，是以枚舉的次數不用很多就可以将命中機率趨近與1。

代碼

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;

struct Point {
    ll x, y;
    Point(ll x = , ll y = ): x(x), y(y) {}
    void read() { scanf("%lld%lld", &x, &y); } 

    bool operator < (const Point& u) const { return x < u.x || (x == u.x && y < u.y); }
    bool operator == (const Point& u) const { return !(*this < u) && !(u < *this); }
    bool operator != (const Point& u) const { return !(*this == u); }
    bool operator > (const Point& u) const { return u < *this; }
    bool operator <= (const Point& u) const { return *this < u || *this == u; }
    bool operator >= (const Point& u) const { return *this > u || *this == u; }
    Point operator + (const Point& u) { return Point(x + u.x, y + u.y); }
    Point operator - (const Point& u) { return Point(x - u.x, y - u.y); }
    Point operator * (const double u) { return Point(x * u, y * u); }
    Point operator / (const double u) { return Point(x / u, y / u); }
    ll operator ^ (const Point& u) { return x*u.y - y*u.x; }
};

typedef Point Vector;
ll Cross(Vector a, Vector b) { return a.x * b.y - a.y * b.x; }

ll getArea (Point* p, int n) {
    ll ret = ;
    for (int i = ; i < n-; i ++)
        ret += Cross(p[i]-p[], p[i+]-p[]);
    return ret <  ? -ret : ret;
}

int getConvexHull (Point* p, int n, Point* ch) {
    sort(p, p + n);

    int m = ;
    for (int i = ; i < n; i++) {
        while (m >  && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
        ch[m++] = p[i];
    }
    int k = m;
    for (int i = n-; i >= ; i--) {
        while (m > k && Cross(ch[m-]-ch[m-], p[i]-ch[m-]) <= ) m--;
        ch[m++] = p[i];
    }
    if (n > ) m--;
    return m;
}

const int maxn = ;

int N, K;
ll dp[maxn][maxn];
Point P[maxn], Q[maxn];

ll solve () {
    ll allarea = getArea(P, N);

    if (K >= N) return allarea;
    ll ret = ;
    bool vis[maxn];
    memset(vis, , sizeof(vis));

    int ti = min( * (N / K), N);
    //int ti = min(10, N);
    for (int t = ; t < ti; t++) {
        int s = rand() % N;
        while (vis[s]) s = rand() % N;
        vis[s] = ;

        memset(dp, , sizeof(dp));
        dp[][] = allarea;

        for (int i = ; i <= N; i++) {
            int u = (i + s) % N;
            ll sum = ;
            for (int j = i-; j >= ; j--) {
                int v = (j + s) % N;
                int p = (v + ) % N;
                ll tmp = ((P[p]-P[u])^(P[v]-P[u]));
                if (tmp < ) tmp = -tmp;
                sum += tmp;
                for (int x = K; x > ; x--)
                    dp[i][x] = max(dp[i][x], dp[j][x-]-sum);
            }
        }
        ret = max(ret, dp[N][K]);
    }
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    srand((int)time(NULL));
    for (int kcas = ; kcas <= cas; kcas++) {
        scanf("%d%d", &N, &K);
        for (int i = ; i < N; i++) Q[i].read();
        N = getConvexHull(Q, N, P);
        printf("Case #%d: %lld\n", kcas, solve());
    }
    return ;
}