1. Abridged Problem Statement  
Given n boundary points (xi, yi), xi strictly increasing, that trace the intersection of Red’s straight-line path from the castle (point 1) to the village (point n) with the terrain profile. Red walks along this path. On terrain‐exposed (above‐ground) slope segments he moves at speed vw; when the path lies strictly inside a mountain (under the terrain), he may tunnel horizontally at speed vc. Compute the minimum total time to travel from the first to the last point along this path.

2. Detailed Editorial  
We number the boundary points from 0 to n–1. Between consecutive boundary points i–1→i, the path lies either on the mountain’s surface or inside rock. Two move‐modes:  
• Surface walking along the sloped segment at speed vw.  
• Horizontal tunneling at speed vc, but only inside a mountain. A horizontal tunnel at height y can start at any boundary point j whose yj ≠ y, provided that the horizontal line y = yi passes through the mountain region between points j and j+1.  

Define dp[i] = minimum time to reach boundary point i. Transitions:  
a) Walk from i–1:  
   dp[i] = dp[i–1] + length(i–1→i)/vw.  

b) Tunnel ending at i from some earlier boundary j < i:  
   • Check if yi lies strictly between yj and yj+1 (i.e. the segment j→j+1 spans height yi).  
   • Compute the intersection point I on segment j→j+1 at height y = yi. Coordinates:  
        if yj+1 ≠ yj,  xi_I = xj + (xj+1–xj)*(yi–yj)/(yj+1–yj);  
        else (flat) xi_I = xj+1.  
   • Walk along slope from boundary j to I at speed vw:  
        d1 = dist((xj, yj),(xi_I, yi))  
   • Tunnel horizontally from I to (xi, yi) at speed vc:  
        d2 = |xi – xi_I|  
   dp[i] = min(dp[i], dp[j] + d1/vw + d2/vc).  
   As soon as yj ≤ yi, further j will lie outside that mountain—break.  

c) By symmetry, one can also consider tunnels starting at i and ending at some future boundary j > i. This updates dp[j] similarly and ensures all horizontal shortcuts are considered in both directions.  

Since each index i only loops backward until the current mountain’s lower boundary, the amortized complexity is O(n²) in the worst case but quite efficient for n ≤ 1000.

3. C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << for pairs and vectors to ease debugging (not critical to logic)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto& x: a) out << x << ' ';
    return out;
}

int n;
long double vw, vc;
vector<pair<long double, long double>> points;

// Read input: n, speeds, then n (x,y) points
void read() {
    cin >> n >> vw >> vc;
    points.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> points[i];
    }
}

void solve() {
    // dp[i]: minimum time to reach boundary point i
    vector<long double> dp(n, 1e18L);
    dp[0] = 0.0L;  // start at point 0 with zero time

    for(int i = 0; i < n; i++) {
        auto [xi, yi] = points[i];

        // 1) Walk along the adjacent slope from i-1 to i
        if(i > 0) {
            auto [xprev, yprev] = points[i - 1];
            long double dx = xi - xprev;
            long double dy = yi - yprev;
            long double dist = sqrt(dx*dx + dy*dy);
            dp[i] = min(dp[i], dp[i - 1] + dist / vw);
        }

        // 2) Consider horizontal tunnels ending at i, from all earlier j < i
        //    but stop when the line y = yi no longer lies inside the mountain.
        for(int j = i - 1; j >= 0; j--) {
            auto [xj, yj] = points[j];
            auto [xj1, yj1] = points[j + 1];

            // Determine the vertical span of segment j→j+1
            long double up   = max(yj, yj1);
            long double down = min(yj, yj1);

            // If the horizontal line at yi cuts this segment
            if(down <= yi && yi <= up) {
                // Find x-coordinate of intersection I
                long double xi_I;
                if(yj1 == yj) {
                    // flat segment, choose its end
                    xi_I = xj1;
                } else {
                    long double slope = (xj1 - xj) / (yj1 - yj);
                    xi_I = xj + slope * (yi - yj);
                }

                // Distance from boundary j to I (walking on slope)
                long double d1 = sqrt((xj - xi_I)*(xj - xi_I) + (yj - yi)*(yj - yi));
                // Horizontal tunnel from I to (xi, yi)
                long double d2 = fabsl(xi - xi_I);

                // Update dp[i]
                dp[i] = min(dp[i],
                    dp[j] + d1 / vw + d2 / vc);
            }

            // Once yj ≤ yi, further j lie outside this mountain—cannot tunnel
            if(yj <= yi) break;
        }

        // 3) Similarly, consider tunnels starting at i and ending at j > i
        for(int j = i + 1; j < n; j++) {
            auto [xj, yj]   = points[j];
            auto [xjm, yjm] = points[j - 1];

            long double up   = max(yj, yjm);
            long double down = min(yj, yjm);

            if(down <= yi && yi <= up) {
                // Intersection with segment (j-1)→j at y = yi
                long double xi_I;
                if(yj == yjm) {
                    xi_I = xjm;
                } else {
                    long double slope = (xj - xjm) / (yj - yjm);
                    xi_I = xjm + slope * (yi - yjm);
                }

                // Distance from I to boundary j (walking on slope)
                long double d1 = sqrt((xj - xi_I)*(xj - xi_I) + (yj - yi)*(yj - yi));
                // Horizontal tunnel from (xi, yi) to I
                long double d2 = fabsl(xi_I - xi);

                // Update dp[j]
                dp[j] = min(dp[j],
                    dp[i] + d1 / vw + d2 / vc);
            }

            if(yj <= yi) break;
        }
    }

    // Answer is dp at last boundary point
    cout << fixed << setprecision(10) << dp[n - 1] << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
import math

def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    n = int(next(it))
    vw = float(next(it))
    vc = float(next(it))
    pts = [(float(next(it)), float(next(it))) for _ in range(n)]

    # dp[i]: min time to reach boundary point i
    dp = [1e18] * n
    dp[0] = 0.0

    for i in range(n):
        xi, yi = pts[i]

        # 1) Walk along adjacent slope
        if i > 0:
            x0, y0 = pts[i-1]
            d = math.hypot(xi - x0, yi - y0)
            dp[i] = min(dp[i], dp[i-1] + d / vw)

        # 2) Tunnel in from any earlier j
        #    Stop when the terrain dips below yi.
        for j in range(i-1, -1, -1):
            xj, yj   = pts[j]
            xj1, yj1 = pts[j+1]
            lo, hi = min(yj, yj1), max(yj, yj1)

            if lo <= yi <= hi:
                # Compute intersection I on segment j→(j+1) at height yi
                if yj1 != yj:
                    slope = (xj1 - xj) / (yj1 - yj)
                    xI = xj + slope*(yi - yj)
                else:
                    xI = xj1
                # Walk from (xj,yj) to I
                d1 = math.hypot(xI - xj, yi - yj)
                # Tunnel from I to (xi, yi)
                d2 = abs(xi - xI)
                dp[i] = min(dp[i], dp[j] + d1/vw + d2/vc)

            # Once yj ≤ yi, no further mountains cover yi from the left
            if yj <= yi:
                break

        # 3) Tunnel out to any future k
        for k in range(i+1, n):
            xk, yk   = pts[k]
            xkm, ykm = pts[k-1]
            lo, hi = min(ykm, yk), max(ykm, yk)

            if lo <= yi <= hi:
                # Intersection I at height yi on segment (k-1)→k
                if yk != ykm:
                    slope = (xk - xkm) / (yk - ykm)
                    xI = xkm + slope*(yi - ykm)
                else:
                    xI = xkm
                # Walk from I to (xk, yk)
                d1 = math.hypot(xk - xI, yk - yi)
                # Tunnel from (xi, yi) to I
                d2 = abs(xI - xi)
                dp[k] = min(dp[k], dp[i] + d1/vw + d2/vc)

            # Once yk ≤ yi, no further coverage
            if yk <= yi:
                break

    # Print result for last point
    print(f"{dp[n-1]:.10f}")

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
• Use DP over the n boundary points: dp[i] = min time to reach point i.  
• Two move‐modes: slope walking (only between adjacent points) at speed vw, and horizontal tunneling inside mountains at speed vc.  
• For each i, update dp[i] via dp[i–1] + slope_length/vw, and via tunnels from any j<i whose segment (j,j+1) spans yi: walk j→intersection/vw then tunnel →i/vc. Similarly propagate tunnels forward.  
• Early breaks when terrain height falls below yi ensure O(n²) works for n≤1000.