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  2. Algorithms
  3. Dynamic Programming
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Road Maintenance

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  • yashparihar729
     + 0 comments

    Here is my solution in java, javascript, python, C, C++, Csharp HackerRank Road Maintenance Problem Solution

  • thecodingsoluti2
     + 0 comments

    Here is Road Maintenance problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-road-maintenance-problem-solution.html

  • TChalla
     + 0 comments
    #include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define foreach(i,x) for(type(x)i = x.begin(); i != x.end(); i++)
#define FOR(ii, aa, bb) for(int ii = aa; ii <= bb; ii++)
#define type(x) __typeof(x.begin())

typedef long long ll;

const int mod = (int) 1e9 + 7;
const int N = 2e5 + 5;

int n, m, x, y, dp[N][11][11], temp[N][11][11];
vector<int> v[N];

void dfs(int node, int root) {
    dp[node][0][0] = 1;
    foreach(it, v[node]) {
        if(*it == root) continue;
        dfs(*it, node);
        FOR(i, 0, 10){
            FOR(j, 0, 10) {
                temp[node][i][j] = dp[node][i][j];
                dp[node][i][j] = 0;
            }
        }
        FOR(i, 0, m){
            FOR(j, 0, m - i){
                FOR(k, 0, m){
                    dp[node][i + j][k] = (dp[node][i + j][k] + temp[node][i][k] * (ll) dp[*it][j][0]) % mod;
                    dp[node][i + j][k + 1] = (dp[node][i + j][k + 1] + temp[node][i][k] * (ll) dp[*it][j][1]) % mod;
                    if(k) dp[node][i + j + 1][k - 1] = (dp[node][i + j + 1][k - 1] + temp[node][i][k] * (ll) dp[*it][j][1] % mod * k) % mod;
                    dp[node][i + j + 1][k] = (dp[node][i + j + 1][k] + temp[node][i][k] * (ll) dp[*it][j][1] % mod) % mod;
                }
            }
        }
    }   
    FOR(i, 0, m) dp[node][i][1] = (dp[node][i][1] + dp[node][i][0]) % mod;
}

int main() {
    cin >> n >> m;
    FOR(i, 2, n) {
        cin >> x >> y;
        v[x].pb(y);
        v[y].pb(x);
    }
    dfs(1, 0);
    cout << dp[1][m][0] << endl;
    return 0;
}

  • nitin29
     + 0 comments

    The following algorithm should work in O(N) complexity. When the road link count is 10000 then the total number of roads is n/2*(n-1) = 49995000. Lets say the road numbers run from 1-49995000 or to n/2*(n-1) . Start navigating the road links. Let’s say there is a link 1-2. Assign this link the next available road number, which is one in this case. So the road lookup will get the following entries 1-(1,2) and reverse. Also the road ending in a node DS will have the following entries 1-null, 2-(1).

    Let’s process link 2-3 this will add one more road, road #2. Now road look up has 2 entries, 1-(1,2), 2-(2,3). The road ending in a node DS will have the following entries 1-null, 2-(1), 3-(2). Since the start of this link node 2 has an entry in road ending in a node DS, so road 1 will be joined with road 2 to form a new road #3. Now road lookup has one more entry 3-(1,3). Also since this road ends in road #3 the corresponding entry in road ending in a node DS is updates to 3-(2,3). Also the road parts DS will have one entry now, 3-(1,2). This DS stores a link for all the single road parts for this road. The entry for road 2 in road ending in a node DS will also get updated 2-(1,2)

    We introduce one more DS here, road grouping. This DS will store the list of all the road with which the given road has at-least one common road, and hence are connected. At this point the DS will have the following entries. 1-(3), 2-(3), 3-(1,2). As you can see the entries in this DS are symmetrical. What this means that road #1 can be paired with 2. So one one pair of disconnected road is possible among the three roads. Lets process link 2-4. This will add one more road road #3. Now road look up has fourth entry 4-(2,4). The road ending in a node DS has one more entry 4-(4). Since the start of this road road #2 has an entry in this DS which point to 2 roads ending in #2 i.e road #1,2. These first roads will combine with road 4 and form a new road, road #5. The road look up will have the new entries as 5-(1,4) and 6-(3,4). The road parts DS will have a new entry for 5-(1,4). The road grouping will be updated with a blank entry for 4-null. The entry for 5 will be 5-(1,3,4), the following entries will be updated to 1-(3,5), 3-(1,2,5),4-(5).

    The second road will combine with 4 to form road #6. The road look up will have the new entry as 6-(3,4). The entry for 6 in road parts will be 6-(2,4). The entry for 6 in road grouping will be 6-(2,3,4,5), the following entries will be updated to 2-(3,6), 3-(1,2,5, 6), 4-(5, 6), 5-(1,3,4,6)

    Following will be the entries in road grouping 1-(3,5), 2-(3,6), 3-(1,2,5,6),4-(5,6),5-(1,3,4,6), (2,3,4,5). So from single road entries of 3, 2 can be selected in 3 ways. Road 3 can paired with 4. Road 5 can be paired with 2. Road 6 can be paired with with road 1. So a total of 6 possibilities.

  • Nagesh
     + 1 comment

    The problem is not clear. Can you please reword or elaborate?

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