# Program to find minimum number of roads we have to make to reach any city from first one in C++

C++Server Side ProgrammingProgramming

Suppose we have two lists costs_from and costs_to of same size where each index i represents a city. It is making a one-way road from city i to j and their costs are costs_from[i] + costs_to[j]. We also have a list of edges where each edge contains [x, y] indicates there is already a one-way road from city x to y. If we want to go to any city from city 0, we have to find the minimum cost to build the necessary roads.

So, if the input is like costs_from = [6, 2, 2, 12] costs_to = [2, 2, 3, 2] edges = [[0, 3]], then the output will be 13, as we can go from 0 to 2 for a cost of 9. After that, we can go from 2 to 1 for a cost of 4. And we already have the road to go to 3 from 0. So total is 9 + 4 = 13.

To solve this, we will follow these steps −

• n := size of costs_from
• ret := 0
• Define two maps edges and redges
• for all item it in g:
• insert it[1] at the end of edges[it[0]]
• insert it[0] at the end of redges[it[1]]
• from_cost := inf
• Define one set visited and another set reachable
• define a function dfs, this will take a number i
• if i is not visited and i is not reachable, then:
• insert i into visited
• for all j in edges[i], do
• dfs(j)
• insert i at the end of po
• define a function dfs2, this will take a number i
• if i is visited, then
• return true
• if i is reachable
• return false
• mark i as visited and mark i as reachable
• ret := true
• for all j in redges[i], do
• ret :+ ret AND dfs2(j)
• return ret
• Define one queue q
• insert 0 into reachable and insert 0 into q
• while (q is not empty), do:
• node := first element of q
• delete element from q
• for each i in edges[node]
• if i is not in reachable, then:
• insert i into reachable, insert i into q
• from_cost := minimum of from_cost and costs_from[node]
• global_min := minimum element of costs_from
• ret := ret + from_cost - global_min
• Define an array po
• for i in range 0 to n, do
• dfs(i)
• reverse the array po
• for each i in po, do
• if i is reachable, then:
• go for next iteration
• clear the visited array
• initial := dfs2(i)
• if initial is true, then:
• best := inf
• for each j in visited:
• best := minimum of best and costs_to[j]
• ret := ret + global_min + best
• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include
using namespace std;
class Solution {
public:
int solve(vector& costs_from, vector& costs_to, vector>& g) {
int n = costs_from.size();
int ret = 0;
map> edges;
map> redges;
for (auto& it : g) {
edges[it[0]].push_back(it[1]);
redges[it[1]].push_back(it[0]);
}
int from_cost = INT_MAX;
set visited;
set reachable;
queue q;
reachable.insert(0);
q.push(0);
while (!q.empty()) {
int node = q.front();
q.pop();
for (int i : edges[node]) {
if (!reachable.count(i)) {
reachable.insert(i);
q.push(i);
}
}
from_cost = min(from_cost, costs_from[node]);
}
int global_min = *min_element(costs_from.begin(), costs_from.end());
ret += from_cost - global_min;
vector po;
function dfs;
dfs = [&](int i) {
if (!visited.count(i) && !reachable.count(i)) {
visited.insert(i);
for (int j : edges[i]) {
dfs(j);
}
po.push_back(i);
}
};
for (int i = 0; i < n; i++) dfs(i);
reverse(po.begin(), po.end());
function dfs2;
dfs2 = [&](int i) {
if (visited.count(i)) return true;
if (reachable.count(i)) return false;
visited.insert(i);
reachable.insert(i);
bool ret = true;
for (int j : redges[i]) {
ret &= dfs2(j);
}
return ret;
};
for (int i : po) {
if (reachable.count(i)) continue;
visited.clear();
bool initial = dfs2(i);
if (initial) {
int best = INT_MAX;
for (int j : visited) {
best = min(best, costs_to[j]);
}
ret += global_min + best;
}
}
return ret;
}
};

int solve(vector& costs_from, vector& costs_to, vector>& edges) {
return (new Solution())->solve(costs_from, costs_to, edges);
}
int main(){
vector costs_from = {6, 2, 2, 12};
vector costs_to = {2, 2, 3, 2};
vector> edges = {{0, 3}};
cout << solve(costs_from, costs_to, edges);
}

## Input

{6, 2, 2, 12}, {2, 2, 3, 2}, {{0, 3}}

## Output

13
Published on 03-Dec-2020 10:10:41