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C++ program to find out the shortest cost path in a given graph for q queries
Suppose, we are given a graph that contains n vertices and is minimally connected. The edges are given to us an array where the edges are given in a {source, dest, weight} format. Now, we are given q number of queries where each query is of the format {source, destination}. We have to find the shortest cost path from the source to the destination via vertex k. We print the cost of the path for each query.
So, if the input is like n = 6, q = 3, k = 1, edges = {{1, 2, 2}, {1, 3, 4}, {3, 4, 2}, {3, 5, 3}, {5, 6, 2}}, queries = {{1, 4}, {2, 6}, {2, 5}}, then the output will be 6 11 9.
Steps
To solve this, we will follow these steps −
Define one 2D array graph of pairs of size n * n Define an array pathTotal of size n Define a function dfs(), this will take a, b, for each value i at graph[a]: if first value of i is same as b, then: Ignore following part, skip to the next iteration pathTotal[first value of i] := pathTotal[a] + second value of i dfs(first value of i, a) for initialize i := 0, when i < n - 1, update (increase i by 1), do: a := first value of (edges[i]) b := second value of (edges[i]) c := third value of (edges[i]) decrease a and b by 1 insert pair (b, c) at the end of graph[a] insert pair (a, c) at the end of graph[b] (decrease k by 1) dfs(k, k) for initialize i := 0, when i < q, update (increase i by 1), do: x := first value of queries[i] y := second value of queries[i] decrease x and y by 1 print(pathTotal[x] + pathTotal[y])
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int,int>>> graph;
vector<int> pathTotal;
int k;
void dfs(int a, int b){
for(auto i : graph.at(a)){
if(i.first == b) continue;
pathTotal.at(i.first) = pathTotal.at(a) + i.second;
dfs(i.first,a);
}
}
void solve(int n, int q, vector<tuple<int, int, int>> edges,
vector<pair<int, int>> queries){
int a, b, c, x, y;
graph.resize(n);
pathTotal.resize(n);
for(int i = 0; i < n - 1; i++){
a = get<0> (edges[i]);
b = get<1> (edges[i]);
c = get<2> (edges[i]);
a--, b--;
graph.at(a).push_back(make_pair(b, c));
graph.at(b).push_back(make_pair(a, c));
}
k--;
dfs(k, k);
for(int i = 0; i < q; i++){
x = queries[i].first;
y = queries[i].second;
x--, y--;
cout << pathTotal.at(x) + pathTotal.at(y) << endl;
}
}
int main() {
int n = 6, q = 3;
k = 1;
vector<tuple<int, int, int>> edges = {{1, 2, 2}, {1, 3, 4}, {3, 4, 2}, {3, 5, 3}, {5, 6, 2}};
vector<pair<int, int>> queries = {{1, 4}, {2, 6}, {2, 5}};
solve(n, q, edges, queries);
return 0;
}Input
6, 3, 1, {{1, 2, 2}, {1, 3, 4}, {3, 4, 2}, {3, 5, 3}, {5, 6, 2}}, {{1, 4}, {2, 6}, {2, 5}}
Output
6 11 9
Updated on: 02-Mar-2022
410 Views
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