Minimum Cost of a Path With Special Roads | MUST DO | TRICKY | OA

https://leetcode.com/problems/minimum-cost-of-a-path-with-special-roads/

You are given an array start where start = [startX, startY] represents your initial position (startX, startY) in a 2D space. You are also given the array target where target = [targetX, targetY] represents your target position (targetX, targetY).
The cost of going from a position (x1, y1) to any other position in the space (x2, y2) is |x2 - x1| + |y2 - y1|.
There are also some special roads. You are given a 2D array specialRoads where specialRoads[i] = [x1i, y1i, x2i, y2i, costi] indicates that the ith special road can take you from (x1i, y1i) to (x2i, y2i) with a cost equal to costi. You can use each special road any number of times.

Return the minimum cost required to go from (startX, startY) to (targetX, targetY).

Example 1:
Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]
Output: 5
Explanation: The optimal path from (1,1) to (4,5) is the following:
- (1,1) -> (1,2). This move has a cost of |1 - 1| + |2 - 1| = 1.
- (1,2) -> (3,3). This move uses the first special edge, the cost is 2.
- (3,3) -> (3,4). This move has a cost of |3 - 3| + |4 - 3| = 1.
- (3,4) -> (4,5). This move uses the second special edge, the cost is 1.
So the total cost is 1 + 2 + 1 + 1 = 5.
It can be shown that we cannot achieve a smaller total cost than 5.

Example 2:
Input: start = [3,2], target = [5,7], specialRoads = [[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]]
Output: 7
Explanation: It is optimal to not use any special edges and go directly from the starting to the ending position with a cost |5 - 3| + |7 - 2| = 7.


Constraints:

start.length == target.length == 2
1 <= startX <= targetX <= 10^5
1 <= startY <= targetY <= 10^5
1 <= specialRoads.length <= 200
specialRoads[i].length == 5
startX <= x1i, x2i <= targetX
startY <= y1i, y2i <= targetY
1 <= costi <= 10^5

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class Solution {
public:
    int minimumCost(vector<int>& start, vector<int>& target, vector<vector<int>>& specialRoads) {
        queue<pair<int,pair<int,int>>> q;

        int n=specialRoads.size();
        vector<int> dp(n+1,INT_MAX);

        q.push({0,{start[0],start[1]}});

        int res=INT_MAX;

        while(!q.empty()){
            auto info=q.front();
            q.pop();

            int dist=info.first;
            int x=info.second.first;
            int y=info.second.second;

            res=min(res,dist+abs(target[0]-x)+abs(target[1]-y));

            int i=0;
            for(auto c: specialRoads){
                int sx=c[0];
                int sy=c[1];
                int dx=c[2];
                int dy=c[3];
                int cost=c[4];
                if(abs(x-sx)+abs(y-sy)+dist+cost<dp[i]){
                    q.push({abs(x-sx)+abs(y-sy)+dist+cost,{dx,dy}});
                    dp[i]=abs(x-sx)+abs(y-sy)+dist+cost;
                }
                i++;
            }
        }
        return res;
    }
};