[ABC130F] Minimum Bounding Box

原题链接

• 题意：给出 (n <= 1e5) 个点在二维平面的坐标，并且给出每个点的运动方向，点不会停止,会一直运动下去，速度为1m/s,然后求过了多少时间，((X_{max}-X_{min})*(Y_{max}-Y_{min})) 的值最小。
• 题解：想到了可能是一个凹函数，但是没想到用三分就做了，还一直死磕规律。
• 代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <cstring>
#include <map>
using namespace std;
const int N = 2e5 + 99;
typedef long long ll;
const ll mod = 1000000007;
const int inf = 0x3f3f3f3f;
struct node {
    string op;
    double x, y;
}a[N];
int n;
double sum(double t) {
    double maxX, minX, maxY, minY;
    for (int i = 1; i <= n; i++) {
        double x = a[i].x;
        double y = a[i].y;
        if (a[i].op == "R")x += t;
        if (a[i].op == "L")x -= t;
        if (a[i].op == "U")y += t;
        if (a[i].op == "D")y -= t;
        if (i == 1) {
            maxX = x;
            minX = x;
            maxY =y;
            minY = y;
        }
        else {
            minX = min(x, minX);
            minY = min(y, minY);
            maxX = max(x, maxX);
            maxY = max(y, maxY);
        }
    }
    //printf("%.10lf %.10lf
", l, r);
    return (maxX - minX) * (maxY - minY);
}
void solve() {
   cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i].x >> a[i].y;
        cin >> a[i].op;
    }
    double l = 0, r = 999999999999.9;
    int TT = 75;
    while (TT--) {
        double lmid = (l + (r - l) / 3);
        double rmid = (l + (r - l) * 2 /3);
        if (sum(lmid) < sum(rmid)) r = rmid;
        else l = lmid;
       //printf("%.10lf %.10lf
", l, r);
    }
    printf("%.14lf
", sum(l));
    // << endl;
}
signed main() {
    ios::sync_with_stdio(0);
    //init();
    ll t = 1;//cin >> t;
    while (t--)solve();
    return 0;
}