题解:P1378 油滴扩展

思路：

1.准备：计算r的最大值

2.搜索：枚举两点间距离，根据前面的油确定当前油的半径 //当两个油滴之间的距离小于那个油滴的半径，我们就要把它变成0

3.答案：原空间减去目前最大空间

 

代码：

#include <iostream>
#include <cmath>
using namespace std;

const int MAXN = 110;
const double pi = acos(-1);

struct node
{
    int x, y;
};

int n, used[MAXN];
double r[MAXN], R[MAXN], ans;
node s, e;
node a[MAXN];

double dis(int x1, int y1, int x2, int y2)
{
    return abs(sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)));
}

void dfs(int id)
{
    if(id > n)
    {
        double sum = 0;
        for(int i=1; i<=n; i++)
        {
            sum += pi * r[i] * r[i];
        }
        ans = max(ans, sum);
        return ;
    }
    for(int i=1; i<=n; i++)
    {
        if(!used[i])
        {
            r[i] = R[i];
            used[i] = 1;
            for(int j=1; j<=n; j++)
            {
                if(used[j] && i != j)
                {
                    double dist = dis(a[j].x, a[j].y, a[i].x, a[i].y);
                    r[i] = min(r[i], dist - r[j]);
                    if(r[i] < 0)
                        r[i] = 0;
                }
            }
            dfs(id + 1);
            used[i] = 0;
        }
    }
}

int main()
{
    cin >> n;
    cin >> s.x >> s.y >> e.x >> e.y;
    for(int i=1; i<=n; i++)
    {
        cin >> a[i].x >> a[i].y;
        R[i] = min(min(abs(s.x - a[i].x), abs(e.x - a[i].x)), min(abs(s.y - a[i].y), abs(e.y - a[i].y)));
    }
    dfs(1);
    double S = (double)(abs(e.x - s.x)) * (double)(abs(e.y - s.y));
    cout << (int)(S - ans + 0.5) << endl;
    return 0;
}