CF 1925D - Good Trip

We have a graph on n children. Only m pairs are marked as friends, and each friendship edge has an initial weight f. The teacher performs exactly k excursions. During each excursion, one unordered pair of children is selected uniformly at random from all possible pairs.

codeforcescompetitive-programmingcombinatoricsdpmathprobabilities

CF 1925D - Good Trip

Rating: 1900
Tags: combinatorics, dp, math, probabilities
Solve time: 2m 9s
Verified: yes

Solution

Problem Understanding

We have a graph on n children. Only m pairs are marked as friends, and each friendship edge has an initial weight f.

The teacher performs exactly k excursions. During each excursion, one unordered pair of children is selected uniformly at random from all possible pairs. The same pair may be selected many times because every selection is independent.

If the selected pair is a friendship edge, two things happen:

First, the current friendship value of that edge contributes to the answer for this excursion.

Second, after the excursion finishes, the friendship value of that edge increases by one, affecting future excursions.

If the selected pair is not a friendship edge, it contributes zero and never changes.

We must compute the expected total contribution over all k excursions. The answer is a rational number. We are asked to output it modulo 10^9+7, which means multiplying the numerator by the modular inverse of the denominator.

The graph itself is sparse. Across all test cases, the total number of friendship edges is at most 10^5, while the total number of excursions is at most 2·10^5. This immediately rules out any simulation over all possible states of friendship values. The number of possible histories grows exponentially with k.

The key challenge is that friendship values evolve over time. A friendship edge becomes more valuable if it has already been selected before. At first glance this creates dependencies between excursions, but expectation allows us to separate the process much more aggressively than a direct probabilistic state simulation would suggest.

Several edge cases deserve attention.

Consider a graph with no friendship edges:

n = 100
m = 0
k = 24

Every selected pair contributes zero. The answer is exactly zero. Any formula that divides by the number of friendship edges or assumes at least one friendship must handle this case.

Consider the smallest nontrivial graph:

n = 2
m = 1
k = 3
f = 5

There is only one possible pair, so it is selected every time. The contributions are:

5 + 6 + 7 = 18

A solution that treats every excursion independently and always uses the original friendship value would incorrectly produce 15.

Another subtle case is when friendship values are different:

n = 3
m = 2
edges:
(1,2) = 100
(2,3) = 1

The expected increase generated by future selections depends only on how often friendship edges are chosen, not on their initial weights. Mixing these two effects incorrectly often leads to double counting.

Approaches

A brute-force approach would model the entire stochastic process.

At each excursion we choose one of the C(n,2) pairs. If it is a friendship edge, its value increases. The state of the process is the vector of current friendship values. After one step there are many possible states, after two steps even more, and after k steps the number of states becomes enormous.

Even if we only track how many times each friendship edge has been selected, the state space size is roughly the number of integer compositions of at most k among m edges, which is completely infeasible.

The brute-force idea is correct because it follows the process exactly, but it becomes impossible long before reaching the problem limits.

The observation that unlocks the solution is linearity of expectation.

Instead of tracking entire states, focus on one excursion at a time.

Let

T = C(n,2)

be the total number of possible pairs.

Suppose an edge starts with friendship value f.

If we are currently at excursion i (0-indexed), the value of that edge equals

f + (# times this edge was selected during previous i excursions)

The expected number of previous selections is easy to compute.

Each excursion chooses this edge with probability

1 / T

independently.

Among the first i excursions, the expected number of times this edge was selected is

i / T

Therefore the expected value of this edge at excursion i is

f + i/T

and the probability that the edge is selected at excursion i is again

1/T.

So the expected contribution of this particular edge during excursion i is

(1/T) · (f + i/T).

Summing over all friendship edges gives

(Σf)/T + m·i/T².

Now sum this over all excursions i = 0 ... k-1.

The first term becomes

k·Σf / T.

The second term becomes

m/T² · Σi
= m/T² · k(k-1)/2.

This is already the final formula.

The entire probabilistic process collapses into two aggregate values:

S = Σf

and

m.

No other information about the graph matters.

Approach Time Complexity Space Complexity Verdict
Brute Force Exponential in k Exponential Too slow
Optimal O(m) per test case O(1) Accepted

Algorithm Walkthrough

  1. Read n, m, and k.
  2. Compute
T = n(n-1)/2

which is the total number of unordered pairs of children. 3. Read all friendship edges and accumulate

S = Σf.

Individual endpoints never appear in the final formula, only the sum of initial friendship values matters. 4. Compute the expected contribution coming from initial friendship values:

k · S / T.

Every excursion selects a particular friendship edge with probability 1/T. 5. Compute the expected contribution generated by friendship-value increases:

m · k(k-1)/2 / T².

For excursion i, every friendship edge has expected extra value i/T, and summing over all edges and all excursions produces this term. 6. Add the two terms modulo 10^9+7. 7. Use modular inverses of T and T² to represent the divisions.

Why it works

For any friendship edge, its current value equals its initial value plus the number of previous times it has been selected.

The number of previous selections is a sum of independent indicator variables. By linearity of expectation, its expectation after i earlier excursions is exactly i/T.

Expectation does not require independence between the edge value and the event that the edge is selected at excursion i, because we directly compute

E[ value × indicator ]

as

(1/T) × E[value].

The indicator for the current excursion is independent of all earlier excursions that determine the current value.

Summing these expectations over all friendship edges and all excursions gives exactly the expected total contribution. Since every transformation preserves equality of expectations, the final formula is correct.

Python Solution

import sys
input = sys.stdin.readline

MOD = 10**9 + 7

def solve():
    t = int(input())

    for _ in range(t):
        n, m, k = map(int, input().split())

        s = 0
        for _ in range(m):
            a, b, f = map(int, input().split())
            s = (s + f) % MOD

        T = n * (n - 1) // 2

        invT = pow(T % MOD, MOD - 2, MOD)
        invT2 = invT * invT % MOD

        part1 = (k % MOD) * s % MOD * invT % MOD

        pairs_sum = (k % MOD) * ((k - 1) % MOD) % MOD
        pairs_sum = pairs_sum * pow(2, MOD - 2, MOD) % MOD

        part2 = (m % MOD) * pairs_sum % MOD * invT2 % MOD

        ans = (part1 + part2) % MOD
        print(ans)

solve()

The implementation follows the derived formula directly.

The variable s stores the sum of all initial friendship values. Since only the sum appears in the final expression, we never need to store individual edges.

T is the number of possible unordered pairs. Division modulo a prime is implemented using modular inverses. Since

T²

appears in the second term, we compute invT once and square it.

The quantity

k(k-1)/2

can be very large, so every multiplication is performed modulo MOD.

A common mistake is to compute the second term using k(k+1)/2. The correct summation is

0 + 1 + ... + (k-1),

which equals k(k-1)/2.

Another common mistake is to forget that only friendship edges can gain value. The factor multiplying the second term is m, not the total number of pairs.

Worked Examples

Example 1

Input:

2 1 10
1 2 1

Here:

T = 1
S = 1
m = 1
k = 10
Variable Value
T 1
S 1
k·S/T 10
k(k-1)/2 45
m·k(k-1)/(2T²) 45
Answer 55

The contributions are actually:

1+2+3+...+10 = 55

and the formula reproduces the exact result.

Example 2

Input:

3 1 2
1 2 1

There are three possible pairs.

Variable Value
T 3
S 1
m 1
k 2
First term 2/3
Second term 1/9
Total 7/9

The expected answer is exactly 7/9, matching the sample.

This example demonstrates the key idea that the growth term comes from expected previous selections. After the first excursion, the friendship edge has probability 1/3 of having increased.

Complexity Analysis

Measure Complexity Explanation
Time O(m) Read friendship edges and accumulate their weights
Space O(1) Only a few running totals are stored

The total number of friendship edges across all test cases is at most 10^5, so the algorithm performs only linear work in the input size. This easily fits within the one second limit.

Test Cases

# helper: run solution on input string, return output string
import sys, io

MOD = 10**9 + 7

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)

    input = sys.stdin.readline
    t = int(input())
    out = []

    for _ in range(t):
        n, m, k = map(int, input().split())

        s = 0
        for _ in range(m):
            a, b, f = map(int, input().split())
            s += f

        T = n * (n - 1) // 2

        invT = pow(T % MOD, MOD - 2, MOD)
        invT2 = invT * invT % MOD

        part1 = (k % MOD) * (s % MOD) % MOD * invT % MOD

        pairs_sum = (k % MOD) * ((k - 1) % MOD) % MOD
        pairs_sum = pairs_sum * pow(2, MOD - 2, MOD) % MOD

        part2 = (m % MOD) * pairs_sum % MOD * invT2 % MOD

        out.append(str((part1 + part2) % MOD))

    return "\n".join(out)

# provided samples
assert run(
"""4
100 0 24
2 1 10
1 2 1
3 1 2
2 1 1
5 2 4
1 2 25
3 2 24
"""
) == """0
55
777777784
40000020"""

# minimum size
assert run(
"""1
2 1 1
1 2 5
"""
) == "5"

# no friendship edges
assert run(
"""1
5 0 100
"""
) == "0"

# single edge, deterministic growth
assert run(
"""1
2 1 3
1 2 5
"""
) == "18"

# multiple edges with equal values
assert run(
"""1
3 3 1
1 2 7
1 3 7
2 3 7
"""
) == "21"
Test input Expected output What it validates
n=2,m=1,k=1 5 Smallest nontrivial case
m=0 0 No friendship edges
n=2,m=1,k=3 18 Repeated growth of the same edge
Complete graph with k=1 21 Growth term disappears when only one excursion exists

Edge Cases

No friendship edges

Input:

1
5 0 10

We have:

S = 0
m = 0

The formula becomes

10·0/T + 0 = 0.

No pair ever contributes anything, so the answer is correctly zero.

Only one possible pair

Input:

1
2 1 4
1 2 3

Here T = 1.

The formula gives:

4·3 + 1·(4·3/2)
= 12 + 6
= 18.

The actual sequence of values is:

3, 4, 5, 6

whose sum is also 18.

Large friendship values

Input:

1
3 1 1
1 2 1000000000

The answer should simply be

1000000000 / 3.

Modulo arithmetic handles this correctly because all divisions are performed using modular inverses. No overflow occurs since Python integers are unbounded and all final computations are reduced modulo 10^9+7.