CF 1245E - Hyakugoku and Ladders

CF 1245E - Hyakugoku and Ladders

Rating: 2300
Tags: dp, probabilities, shortest paths
Solve time: 3m 29s
Verified: no

Solution

Problem Understanding

The game can be viewed as a directed line of states arranged along a fixed serpentine path on a 10 by 10 grid. Each cell corresponds to a position on this path, starting from the bottom-left cell and ending at the top-left cell. Movement along the path is deterministic: within a row you move horizontally in alternating directions, and at row boundaries you move vertically upward. So every cell has a unique successor unless it is the final goal cell.

Each turn consists of rolling a fair six-sided dice. The player attempts to move forward along the path by exactly that many steps. If the dice value would overshoot the goal, the player stays in place. After landing, if the cell contains the base of a ladder, the player may choose to either stay or instantly teleport to the ladder’s endpoint higher up the board. Once a ladder is taken, it cannot be chained immediately even if another ladder starts at the landing cell.

The task is to compute the minimum expected number of turns needed to reach the goal, assuming optimal decisions about whether to use a ladder at each opportunity.

The state space is small and fixed: there are exactly 100 cells. This immediately suggests that any solution with a cubic or worse dependence on the number of states is acceptable, while anything exponential or involving repeated recomputation over dice sequences would be unnecessary. A linear system or shortest-path style dynamic program is feasible.

A subtle issue appears with ladders that land on cells that themselves contain ladders. A naive greedy assumption that “always take the ladder if it goes up” is wrong because taking a ladder may skip a better expected-position cell that offers more favorable future dice transitions. Another failure case is overshooting near the goal: moving 5 or 6 steps from a near-terminal state can lead to the same “no movement” outcome, which affects transition probabilities and must be handled explicitly.

Approaches

A brute-force interpretation treats every game state as a position on the board and every dice outcome as a transition. From each cell we simulate all possible dice rolls, apply movement, optionally choose ladder usage, and recursively compute expected time to finish.

This naive recursion builds a Markov decision process. While correct conceptually, recomputing expected values repeatedly for each state leads to exponential blowup unless memoized. Even with memoization, if transitions are evaluated in a cyclic dependency order, naive recursion may require repeated relaxation until convergence.

The key observation is that this is a finite Markov decision process with non-negative expected costs and monotone progress toward the goal. Each state has a fixed expected value equation that depends only on already defined states or itself, so the system can be solved using iterative relaxation similar to value iteration, or more directly as a shortest-path-like DP on expected cost.

Each state corresponds to a linear equation:

Expected cost at state i equals 1 (for the current turn) plus the average over six dice outcomes of the best successor state. The “best successor” depends on whether we choose to climb a ladder when available. That introduces a local min inside the transition definition.

Thus each state can be computed if all successor states are known. Since movement always goes forward or stays, states can be processed in reverse order from goal to start. This produces a stable DP ordering.

Approach	Time Complexity	Space Complexity	Verdict
Brute force recursion	exponential	O(n)	Too slow
DP over states (reverse order)	O(100 × 6)	O(100)	Accepted

Algorithm Walkthrough

We first flatten the 10 by 10 board into a single array of 100 positions following the serpentine traversal order. Each position i stores either a ladder destination or none.

We then compute the next positions for all dice outcomes. From a state i, for each dice roll r, we determine the target position j by moving r steps forward along the path, clamping at the goal if we overshoot.

Once we reach j, we may either stay at j or, if a ladder starts there, jump to its endpoint. We choose whichever option yields smaller expected time.

We define dp[i] as the minimum expected number of turns required to reach the goal from position i.

We process positions in reverse order from 99 down to 0.

1. Initialize dp at the goal cell as 0 because no turns are needed once the game is finished.
2. For each other position i, compute the expected value of taking a turn. Each turn always costs 1.
3. For each dice outcome r from 1 to 6, compute the landing cell j after moving r steps, or the goal if movement overshoots. This gives a deterministic next position before ladder decisions.
4. If j has a ladder, compute two candidate outcomes: staying at j and taking the ladder destination. Since the player acts optimally, choose the minimum expected dp value between these two states.
5. Average these six resulting dp values and add 1 to account for the current turn. Assign this to dp[i].
6. Continue until all states are processed, ensuring that any state’s dependencies (all forward positions) have already been computed.

The core reason this ordering works is that movement never decreases the index along the path. Every transition goes strictly forward or remains at the goal, so dp depends only on higher-index states. This makes the recurrence acyclic and guarantees that a single backward pass is sufficient.

Python Solution

import sys
input = sys.stdin.readline

n = 10
grid = [list(map(int, input().split())) for _ in range(n)]

# flatten board in serpentine order
pos = []
coords = []
for i in range(n):
    row = n - 1 - i
    if i % 2 == 0:
        for col in range(n):
            pos.append((row, col))
            coords.append((row, col))
    else:
        for col in range(n - 1, -1, -1):
            pos.append((row, col))
            coords.append((row, col))

idx = {coords[i]: i for i in range(100)}

ladder = [0] * 100
for i in range(n):
    for j in range(n):
        if grid[i][j] > 0:
            start = idx[(n - 1 - i, j)]
            ladder[start] = start + grid[i][j] * 10

def clamp(x):
    return min(x, 99)

dp = [0.0] * 100
dp[99] = 0.0

for i in range(98, -1, -1):
    total = 0.0
    for dice in range(1, 7):
        j = i + dice
        if j >= 100:
            j = 99
        best = dp[j]
        if ladder[j]:
            best = min(best, dp[ladder[j]])
        total += best
    dp[i] = 1.0 + total / 6.0

print(dp[0])

The code first constructs the exact path ordering of all 100 squares so that each move becomes a simple index increment. The ladder array maps each starting cell to its destination index in this linearization.

The DP array stores expected values, and we iterate from the end backward because all transitions go forward in this index space. For each dice roll, we compute the landing index and then apply the optimal ladder decision locally by comparing “stay” versus “climb”.

The final answer is dp[0], corresponding to the starting cell.

Worked Examples

We use the no-ladder case since it isolates the probabilistic structure.

Example: empty board

State i	dp[i] computation
99	0
98	1 + average(dp[99] over 6 rolls) = 1
97	1 + (dp[98]+dp[99] mix)/6
...	...
0	accumulates full expectation

At each step, the value represents the expected number of turns needed when only deterministic forward movement exists. Near the end, many dice outcomes saturate at the goal, causing repeated dp[99] contributions of 0.

This demonstrates that overshooting is naturally handled by clamping transitions to the goal state, which reduces expected value growth near the end.

Example: single ladder near start

Suppose a ladder exists at position 2 that jumps to position 20.

When computing dp[2], we compare two options for transitions landing at 2: staying contributes dp[2], while climbing contributes dp[20]. If dp[20] is smaller, the DP will prefer climbing; otherwise it will ignore the ladder.

This shows that ladder decisions are local optimizations embedded inside global expectation, and they do not require expanding the state space.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(100 × 6)	each state evaluates 6 dice outcomes once
Space	O(100)	DP array plus linearized board

The state space is constant size, so the solution runs in constant time in practice and easily satisfies limits. Even if generalized to larger grids, the same structure would remain linear in number of states times dice outcomes.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import math
    n = 10
    grid = [list(map(int, sys.stdin.readline().split())) for _ in range(n)]

    pos = []
    coords = []
    for i in range(n):
        row = n - 1 - i
        if i % 2 == 0:
            for col in range(n):
                coords.append((row, col))
        else:
            for col in range(n - 1, -1, -1):
                coords.append((row, col))

    idx = {coords[i]: i for i in range(100)}
    ladder = [0] * 100

    for i in range(n):
        for j in range(n):
            if grid[i][j] > 0:
                start = idx[(n - 1 - i, j)]
                ladder[start] = start + grid[i][j] * 10

    dp = [0.0] * 100
    for i in range(98, -1, -1):
        s = 0.0
        for d in range(1, 7):
            j = min(99, i + d)
            best = dp[j]
            if ladder[j]:
                best = min(best, dp[ladder[j]])
            s += best
        dp[i] = 1 + s / 6.0

    return str(dp[0])

# provided sample
assert abs(float(run(
"""0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
""").strip()) - 33.0476190476) < 1e-6

Test input	Expected output	What it validates
Empty board	33.0476	baseline probability-only DP
Single ladder	lower than no-ladder	ladder decision correctness
Ladder to goal	immediate shortcut effect	terminal transition handling
All ladders zero except start	normal movement only	boundary correctness

Edge Cases

A ladder that leads directly to the goal tests whether the DP correctly treats the goal as absorbing. From that landing cell, dp is zero, so taking the ladder must always dominate unless it causes no benefit.

Overshooting the goal with dice rolls ensures that all values greater than remaining distance map correctly to the goal state. This prevents array overflows and ensures the expectation does not incorrectly propagate beyond bounds.

Multiple ladders landing on the same cell are handled safely because the DP only compares destination values, not ladder identity, so convergence is unaffected.