洛谷1378 油滴扩展

原题链接

爆搜水题，数据小，连剪枝也不用。
对于每个油滴，扩展的最大半径即是对圆心到\(x\)边界的距离、圆心到\(y\)边界的距离、与每个已经扩展的油滴圆心距离减去该油滴扩展的半径 取最小值，注意对\(0\)取\(max\)。

#include<cstdio>
#include<cmath>
using namespace std;
const int N = 10;
const double pi = 3.14159265358979323846;
struct dd{
    int x, y, pre, suc;
};
dd a[N];
int sta[N], sqlx, sqly, sqrx, sqry, n, tp;
double dis[N][N], edg[N], R[N], ma;
inline int re()
{
    int x = 0;
    char c = getchar();
    bool p = 0;
    for (; c < '0' || c > '9'; c = getchar())
        p |= c == '-';
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return p ? -x : x;
}
inline double minn(double x, double y)
{
    return x < y ? x : y;
}
inline double maxn(double x, double y)
{
    return x > y ? x : y;
}
inline double js(double x)
{
    return x < 0 ? -x : x;
}
inline void de(int x)
{
    a[a[x].pre].suc = a[x].suc;
    a[a[x].suc].pre = a[x].pre;
}
inline void con(int x)
{
    a[a[x].pre].suc = x;
    a[a[x].suc].pre = x;
}
void dfs(int x, double s)
{
    if (!(x ^ n))
        ma = maxn(ma, s);
    int i, j;
    double r;
    for (i = a[0].suc; i <= n; i = a[i].suc)
    {
        r = edg[i];
        for (j = 1; j <= tp; j++)
            r = minn(r, maxn(dis[i][sta[j]] - R[j], 0));
        sta[++tp] = i;
        R[tp] = r;
        de(i);
        dfs(x + 1, s + pi * r * r);
        con(i);
        tp--;
    }
}
int main()
{
    int i, j;
    n = re();
    sqlx = re();
    sqly = re();
    sqrx = re();
    sqry = re();
    for (i = 1, a[0].suc = 1, a[n + 1].pre = n; i <= n; i++)
    {
        a[i].x = re();
        a[i].y = re();
        a[i].pre = i - 1;
        a[i].suc = i + 1;
    }
    for (i = 1; i < n; i++)
        for (j = i + 1; j <= n; j++)
            dis[i][j] = dis[j][i] = sqrt(1.0 * (a[i].x - a[j].x) * (a[i].x - a[j].x) + 1.0 * (a[i].y - a[j].y) * (a[i].y - a[j].y));
    for (i = 1; i <= n; i++)
        edg[i] = minn(minn(js(a[i].x - sqlx), js(a[i].x - sqrx)), minn(js(a[i].y - sqly), js(a[i].y - sqry)));
    dfs(0, 0);
    printf("%.0f", js(sqlx - sqrx) * js(sqly - sqry) - ma);  
    return 0;
}