CodeForces 852 H. Bob and stages

链接：

link

题意：

给平面上n个点，选出k个组成一个凸包，同时要求凸包内部没有点，求最大面积。

题解：

首先枚举左下角的点，然后极角排序，处理出所有可能的转移。

转移边是个星型图，大概就按照极角序每个点维护一个队列表示用来维护未来可能存在的转移。

然后记f(i,j,k)表示考虑最后一条转移是i→j，选了k条边的最大面积，转移到f(l,i,k+1)，用前缀和优化一下，最终复杂度可以做到O(n3k)。

可能题解写的不是很清楚，详见代码。

代码：

#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define mset(x, y) memset(x, y, sizeof x)
#define mcpy(x, y) memcpy(x, y, sizeof x)
using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

inline int Read()
{
int x = 0, f = 1, c = getchar();
for (; !isdigit(c); c = getchar())
if (c == '-')
f = -1;
for (;  isdigit(c); c = getchar())
x = x * 10 + c - '0';
return x * f;
}

const int MAXN = 205;
const int MAXM = 55;

struct Point
{
int x, y;
double k;

Point(int _x = 0, int _y = 0)
{
x = _x, y = _y;
}

Point operator - (const Point &b) const
{
return Point(x - b.x, y - b.y);
}

bool operator < (const Point &b) const
{
return k < b.k;
}

LL operator * (const Point &b) const
{
return 1LL * x * b.y - 1LL * y * b.x;
}
} a[MAXN], p[MAXN];

int n, m;
LL ans, mx[MAXM], f[MAXN][MAXN][MAXM];
vector <int> adj[MAXN], rev[MAXN];
queue <int> q[MAXN];

inline bool Onleft(Point a, Point b, Point c)
{
return (b - a) * (c - b) > 0;
}

inline void Add(int x, int y)
{
while (!q[x].empty() && Onleft(p[q[x].front()], p[x], p[y]))
Add(q[x].front(), y), q[x].pop();
rev[x].pb(y);
adj[y].pb(x);
q[y].push(x);
}

inline void Init(int n)
{
for (int i = 1; i <= n; i ++)
{
adj[i].clear(), rev[i].clear();
while (!q[i].empty())
q[i].pop();
}
for (int i = 1; i < n; i ++)
Add(i, i + 1);
}

inline void Dp(int n)
{
mset(f, 0);
for (int i = n; i; i --)
{
for (int j = 0; j < m; j ++)
mx[j] = 0;
reverse(adj[i].begin(), adj[i].end());
int cur = rev[i].size() - 1;
for (auto j : adj[i])
{
f[j][i][1] = p[j] * p[i];
for (int k = 2; k <= m; k ++)
if (mx[k - 1])
f[j][i][k] = mx[k - 1] + p[j] * p[i];
for (; ~cur && Onleft(p[j], p[i], p[rev[i][cur]]); cur --)
for (int k = 1; k < m; k ++)
if (mx[k] < f[i][rev[i][cur]][k])
mx[k] = f[i][rev[i][cur]][k], f[j][i][k + 1] = mx[k] + p[j] * p[i];
}
}
for (int i = 1; i <= n; i ++)
for (auto j : rev[i])
ans = max(ans, f[i][j][m - 2]);
}

inline void Solve(int idx)
{
int cnt = 0;
for (int i = 1; i <= n; i ++)
if (i ^ idx)
if (a[i].x > a[idx].x || (a[i].x == a[idx].x && a[i].y > a[idx].y))
p[++ cnt] = a[i] - a[idx], p[cnt].k = atan2(p[cnt].y, p[cnt].x);
sort(p + 1, p + cnt + 1);
Init(cnt);
Dp(cnt);
}

int main()
{
#ifdef wxh010910
freopen("data.in", "r", stdin);
#endif
n = Read(), m = Read();
for (int i = 1; i <= n; i ++)
a[i].x = Read(), a[i].y = Read();
for (int i = 1; i <= n; i ++)
Solve(i);
cout << ans / 2 << "." << (ans & 1 ? "50" : "00") << endl;
return 0;
}