131 lines
4.7 KiB
Markdown
131 lines
4.7 KiB
Markdown
# Silver - Graphs
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Author: Siyong Huang
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## Overview
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- Prerequisites
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- Depth First Search (DFS)
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- Flood Fill
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- Graph Two-Coloring
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- Cycle Detection
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## Prerequisites
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- [Graph Theory](https://csacademy.com/lesson/introduction_to_graphs)
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- [Graph Representations](https://csacademy.com/lesson/graph_representation)
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- Note: DFS is most commonly implemented with adjacency lists
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## Depth First Search (DFS)
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*Depth First Search*, more commonly DFS, is a fundamental graph algorithm that traverses an entire connected component. The rest of this document describe various applications of DFS.
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### Tutorial
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- Recommended:
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- [CSAcademy BFS](https://csacademy.com/lesson/depth_first_search/)
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- Additional:
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- CPH Chapter 12
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- [cp-algo DFS](https://cp-algorithms.com/graph/depth-first-search.html)
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### Problems
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- [Mootube, Silver (Easy)](http://usaco.org/index.php?page=viewproblem2&cpid=788)
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- [Closing the Barn, Silver (Easy)](http://usaco.org/index.php?page=viewproblem2&cpid=644)
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- [Moocast, Silver (Easy)](http://usaco.org/index.php?page=viewproblem2&cpid=668)
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- [Pails (Normal)](http://usaco.org/index.php?page=viewproblem2&cpid=620)
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- [Milk Visits (Normal)](http://www.usaco.org/index.php?page=viewproblem2&cpid=968)
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## Flood Fill
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*Flood Fill* refers to finding the number of connected components in a graph, frequently on a grid.
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### Tutorial
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- Recommended:
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- Ch 10 of https://www.overleaf.com/project/5e73f65cde1d010001224d8a
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### Problems
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- [Ice Perimeter (Easy)](http://usaco.org/index.php?page=viewproblem2&cpid=895)
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- [Switching on the Lights (Moderate)](http://www.usaco.org/index.php?page=viewproblem2&cpid=570)
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- [Build Gates (Moderate)](http://www.usaco.org/index.php?page=viewproblem2&cpid=596)
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- [Why Did the Cow Cross the Road III, Silver (Moderate)](http://usaco.org/index.php?page=viewproblem2&cpid=716)
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- [Multiplayer Moo (Hard)](http://usaco.org/index.php?page=viewproblem2&cpid=836)
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## Graph Two-Coloring
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*Graph two-colorings* is assigning a boolean value to each node of the graph, dictated by the edge configuration
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The most common example of a two-colored graph is a *bipartite graph*, in which each edge connects two nodes of opposite colors
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### Tutorial
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The idea is that we can arbitrarily label a node and then run DFS. Every time we visit a new (unvisited) node, we set its color based on the edge rule. When we visit a previously visited node, check to see whether its color matches the edge rule. For example, an implementation of coloring a bipartite graph is shown below.
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```
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bool is_bipartite = true;
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void dfs(int node)
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{
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visited[node] = true;
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for(int u:adj_list[node])
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if(visited[u])
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{
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if(color[u] == color[node])
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is_bipartite = false;
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}
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else
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{
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color[u] = !color[node];
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dfs(u);
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}
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}
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```
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- Additional:
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- [Bipartite Graphs: cp-alg bipartite check](https://cp-algorithms.com/graph/bipartite-check.html)
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- Note: CP Algorithm uses bfs, but dfs accomplishes the same task
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### Problems
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- [The Great Revegetation (Normal)](http://usaco.org/index.php?page=viewproblem2&cpid=920)
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## Cycle Detection
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A *cycle* is a non-empty path of distinct edges that start and end at the same node.
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*Cycle detection* determines properties of cycles in a graph, such as counting the number of cycles in a graph or determining whether a node is in a cycle. For most silver-level cycle problems, each node has only one out-degree, meaning that it's adjacency list is of size 1. If this is not the case, the problem generalizes to *Strongly Connected Components*, a platinum level concept.
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### Tutorial
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The following sample code counts the number of cycles in a graph where each node points to one other node. The "stack" contains nodes that can reach the current node. If the current node points to a node v on the stack (on_stack[v] is true), then we know that a cycle has been created. However, if the current node points to a node v that has been previously visited but is not on the stack, then we know that the current chain of nodes points into a cycle that has already been considered.
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```
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//Each node points to next_node[node]
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bool visited[MAXN], on_stack[MAXN];
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int number_of_cycles = 0;
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void dfs(int n)
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{
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visited[n] = on_stack[n] = true;
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int u = next_node[n];
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if(on_stack[u])
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number_of_cycles++;
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else if(!visited[u])
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dfs(u);
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on_stack[n] = false;
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}
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int main()
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{
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//read input, etc
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for(int i = 1;i <= N;i++)
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if(!visited[i])
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dfs(i);
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}
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```
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### Problems
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- [Codeforces 1020B. Badge (Very Easy)](https://codeforces.com/contest/1020/problem/B)
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- Try to solve the problem in O(N)!
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- [The Bovine Shuffle (Normal)](http://usaco.org/index.php?page=viewproblem2&cpid=764)
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- [Swapity Swapity Swap (Very Hard)](http://www.usaco.org/index.php?page=viewproblem2&cpid=1014)
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