Codeforces Round 365 Div2 Problem C

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CF703C Chris and Road

Solution Sketch

Key idea

Think of the slope of the line $m = \frac{u}{v}$. What can you do with it?

Observation

If you draw the polygon out on the Cartesian coordinate plane and draw a line $y = mx$, you will figure out that if the line never intersects with the polygon, you won’t be hit by the bus!

Otherwise, find the rightmost line using the same slope ($m = \frac{u}{v}$) by iterating over all vertices, and use the value $b$ from the rightmost line ($y = mx + b$) to calculate the minimal time to walk to the target, which is $ \frac{(w - b)}{u} $.

AC Code

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#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> ii;
#define EPS 1e-9
int main()
{
    int n, w, v, u;
    scanf("%d %d %d %d", &n, &w, &v, &u);
    vector<ii> inp;
    double m = (double)u / v;
    for (int i = 0; i < n; i++) {
        int x, y;
        scanf("%d %d", &x, &y);
        inp.push_back(ii(x, y));
    }
    int wrongSide = 0;
    for (int i = 0; i < n; i++) {
        if (m * inp[i].first - inp[i].second > -EPS)
            continue;
        else
            wrongSide++;
    }
    if (wrongSide == 0 || wrongSide == n) {
        printf("%.15f\n", (double)w / u);
    } else {
        double b = INT_MAX;
        for (int i = 0; i < n; i++) {
            b = min(b, inp[i].second - inp[i].first * m);
        }
        printf("%.15f\n", fabs((w - b) / u));
    }
    return 0;
}