Shortest Path? Toph Criterion 2021 Round 9 Explanation with code

Problem link

First of all we need to find the shortest paths from a source node S to destination node T of the given graph.

And secondly a node will be given say X and we need to find the minimum distance from node X to any node of the shortest paths.

For find that a node belongs to the shortest path or not? here is the observations,

we will run a bfs from source node and find the distances of all node from source. And again we will run another bfs from destination node and find the distance of all node from destination. If you have any doubt on bfs check the link

A node will be belong to any shortest path of the graph if distance from source + distance from destination of the node equal to the shortest path distance, in another word,

source_distance[i]+destination_distance[i] == source_distance[destination]

Now we had to find distance from any shortest path to the asking node. We can precalculate all the distances by running a bfs one more time. In this case we will push the nodes into the queue which are belong to the shortest paths and initialize the distance of these nodes as 0. This is also called Multisource bfs .

Code:

C - Digital Graffiti Atcoder Beginner Contest 191 Code with Explanation

Problem Link

Explanation: We need to find number of sides, if  all '#' creates a polygon itself. For this purpose we need to find number of sides of this shape. Lets find if a cell is a side or not.

    1    2    3    4
1    .    .    .    .

2    .    #    #    .

3    .    #    #    .

4    .    .    .    .

Observations: If 4 adjacent cells has odd number of '#' it will contain a side of polygon. If a cell is (i,j) we will count '#' in cells (i,j),(i+1,j),(i+1,j+1),(i,j+1) for cell i,j.

for (1,1) >> Number of #  is 1 at(2,2)

for(1,2)>> Number of # is 2 at(2,2 and 2,3)

for(1,3)>> Number of # is 1 at(2,3)

...

...

...

In this way we will get the number of sides of the polygon.

Code

void solution(){
    int h,w;
    cin>>h>>w;
    string s[h];
    for(int i=0;i<h;i++){
        cin>>s[i];
    }
    int ans=0;
    for(int i=0;i<h-1;i++){
        for(int j=0;j<w-1;j++){
            int cnt=0;
            if(s[i][j]=='#')cnt++;
            if(s[i+1][j]=='#')cnt++;
            if(s[i+1][j+1]=='#')cnt++;
            if(s[i][j+1]=='#')cnt++;
            if(cnt%2!=0)ans++;
        }
    }
    cout << ans << endl;
}