洛谷P1378
这道题需要处理的信息比较多,需要注意的是一个油滴扩展后可能会包含其他的点
#include <iostream>
#include <utility>
#include <vector>
#include <cmath>
using namespace std;
typedef long long ll;
#define fi(i, a, b) for (int i = a; i <= b; ++i)
#define fr(i, a, b) for (int i = a; i >= b; --i)
#define x first
#define y second
#define sz(x) ((int)(x).size())
#define pb push_back
using pii = pair<int, int>;
vector<pair<int, int>> vec;
int up, down, lef, rig;
int full_a[6];
bool vis[6];
double dis[6][6];
double cir[6];
int n;
double maxx;
double ans;
//#define DEBUG
void dfs(int x)
{
if (x == n)
{
ans = 0;
fi(i, 0, n - 1)
{
if(cir[i] == -1)
continue;
else
ans += pow(cir[i], 2);
}
ans = ans * M_PI;
maxx = max(ans, maxx);
return;
}
fi(i, 0, n - 1)
{
if (!vis[i])
{
vis[i] = true;
full_a[x] = i;
double p = min(abs(vec[i].x - lef), abs(vec[i].x - rig));
double q = min(abs(vec[i].y - up), abs(vec[i].y - down));
double minn = min(q, p);
fi(j, 0, n - 1)
{
if (cir[j])
{
minn = min(minn, dis[i][j] - cir[j]);
}
}
if(minn < 0)
cir[i] = -1;
else
cir[i] = minn;
// cout << cir[i] << endl;
dfs(x + 1);
cir[i] = 0;
vis[i] = false;
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
int temp = n;
int x, y, x1, y1;
cin >> x >> y >> x1 >> y1;
up = y, down = y1, lef = x, rig = x1;
while (temp--)
{
int a, b;
cin >> a >> b;
vec.push_back({a, b});
}
fi(i, 0, sz(vec) - 1) fi(j, i + 1, sz(vec) - 1)
{
dis[i][j] = dis[j][i] = sqrt(pow((vec[i].x - vec[j].x), 2) + pow((vec[i].y - vec[j].y), 2));
// cout<< dis[i][j] << endl;
}
dfs(0);
int cnt = abs(up - down) * abs(lef - rig);
// cout << cnt << endl;
cnt = cnt - maxx + 0.5;
cout << cnt << endl;
#ifdef DEBUG
//freopen(D:\in.txt,r,stdin);
#endif
return 0;
}