系统架构评估_3.ATAM方法

2024蓝桥杯赛前模版突击：图论篇

图论在蓝桥杯中一般考的不难，如果有图论的题，就基本是模板题，知道板子就有分了。

邻接表

本文使用方法1的方式实现邻接表

邻接表1

static int[] dist = new int[N],st = new int[N];
static int[] h = new int[N],e = new int[M],ne = new int[M],w = new int[M];
static int idx;

static void init(){
	Arrays.fill(h,-1);
}
static void add(int a,int b,int c) {
	e[idx] = b;
	w[idx] = c;
	ne[idx] = h[a];
	h[a] = idx++;
}

邻接表2

用来快速得到顶点的所有邻边条数

ArrayList<Integer>[] g = new ArrayList[N];

//初始化
for(int i=0;i<n;i++)
    g[i] = new ArrayList<Integer>();
    
//顶点a,b中间添加一条边
g[a].add(b);

最短路Dijkstra

单源最短路 O(mlogn)

package _00模板;
import java.util.*;

public class Dijkstra {
	static int INF = 0x3f3f3f3f;
	static int N = 101000,M = 2*N;
	static int[] dist = new int[N],st = new int[N];
	static int[] h = new int[N],e = new int[M],ne = new int[M],w = new int[M];
	static int idx;
	static int n,m;
	static long ans;
	static void add(int a,int b,int c) {
		e[idx] = b;
		w[idx] = c;
		ne[idx] = h[a];
		h[a] = idx++;
	}
	static int dijkstra(int start) {
		Arrays.fill(dist,INF);
		PriorityQueue<PII> q = new PriorityQueue<>((a,b)->a.dist-b.dist);
		q.add(new PII(start,0));
		st[start] = 0;
		while(q.size()>0) {
			PII top = q.poll();
			if (st[top.v]==1) continue;
			st[top.v] = 1;
			for(int i=h[top.v];i!=-1;i=ne[i]) {
				int j = e[i],val = w[i];
				if(dist[top.v]+val<dist[j]) {
					dist[j] = dist[top.v]+val;
					q.add(new PII(j,dist[j]));
				}
			}
		}
		return dist[n]!=INF?dist[n]:-1;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		Arrays.fill(h,-1);
		n = sc.nextInt();
		m = sc.nextInt();
		for(int i=1;i<=n;i++) {
			
		}
	}
	

}
class PII{
	int dist;int v;
	public PII(int v,int dist) {
		// TODO Auto-generated constructor stub
		this.v = v;
		this.dist = dist;
	}
}

最短路spfa

负权图的最短路O（m*n）

package _00模板;
import java.util.*;
public class Spfa {
	static int INF = 0x3f3f3f3f;
	static int N = 101000,M = 2*N;
	static int[] st = new int[N],dist = new int[N];
	static int n,m;
	static long ans;
	static int[] h = new int[N],e = new int[M],w = new int[M],ne = new int[M];
	static int idx;
	static void add(int a,int b,int c) {
		e[idx] = b;
		w[idx] = c;
		ne[idx] = h[a];
		h[a] = idx++;
	}
	static int spfa(int start) {
		Arrays.fill(dist,INF);
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(start);
		st[start] = 1;
		dist[start] = 0;
		
		while(q.size()>0) {
			int top = q.poll();
			st[top] = 0;
			for(int i=h[top];i!=-1;i=ne[i]) {
				int j = e[i];
				if(dist[top]+w[i]< dist[j]) {
					dist[j] = dist[top]+w[i];
					if(st[j]==0) {
						st[j] = 1;
						q.add(j);
					}	
				}
			}
		}
		return dist[n]!=INF?dist[N]:-1;
		
	}
	
}

Floyd

多源最短路O（n^3）

package _00模板;
import java.util.*;

public class Floyd {
	static int INF = 0x3f3f3f3f;
	static int N = 101000,M = 2*N;
	static int[][] g = new int[N][N];
	static int n,m;
	static long ans;
	static void floyd() {
		for(int k=1;k<=n;k++) {
			for(int i=1;i<=n;i++) {
				for(int j=1;j<=n;j++) {
					g[i][j] = Math.min(g[i][j],g[i][k]+g[k][j]);
				}
			}
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		for(int i=1;i<=n;i++) {
			Arrays.fill(g[i],INF);
			g[i][i] = 0;
		}
		n = sc.nextInt();
		m = sc.nextInt();
		for(int i=1;i<=m;i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int c = sc.nextInt();
			g[a][b] = c;
			g[b][a] = c;
		}
		floyd();
	}
}

最小生成树kruskal

kruskal 算法O (mlogm)，Prim不需要掌握，用kruskal 就行

package _00模板;
import java.util.*;

public class Kruskal {
	static int INF = 0x3f3f3f3f;
	static int N = 101000,M = 2*N;
	static Edge[] edges = new Edge[N];
	static int idx;
	static int n,m;
	static long ans;
	static int[] fa = new int[N];
	static void init() {
		for(int i=1;i<=n;i++) {
			fa[i] = i;
		}
	}
	static int find(int x) {
		if(fa[x]==x) return x;
		return fa[x] = find(fa[x]);
	}
	static void union(int a,int b) {
		fa[find(a)] = find(b);
	}
	static int kruskal() {
		Arrays.sort(edges,0,idx,(a,b)->(a.w-b.w));
		int cnt = 0,res = 0;
		for(int i=0;i<m;i++) {
			int a = edges[i].a;
			int b = edges[i].b;
			int w = edges[i].w;
			if(find(a)!=find(b)) {
				union(a,b);
				cnt += 1;
				res += w;
			}
		}
		return cnt==n-1?res:-1;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		n = sc.nextInt();
		m = sc.nextInt();
		for(int i=1;i<=m;i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int w = sc.nextInt();
			edges[idx++] = new Edge(a,b,w);
		}
	}
}
class Edge{
	int a,b,w;
	public Edge(int a,int b,int w) {
		// TODO Auto-generated constructor stub
		this.a = a;
		this.b = b;
		this.w = w;
	}
}

拓扑排序

	int[] d = new int[N];//存放入度
	int[] print = new int[N];//记录答案
	int cnt;
	
	static boolean topSort() {
		Queue<Integer> q = new LinkedList<Integer>();
		for(int i=1;i<=n;i++) {
			if(d[i]==0)
				q.add(i);
		}
		while(q.size()>0) {
			Integer top = q.poll();
			print[cnt++] = top;
			for(int i=h[top];i!=-1;i=ne[i]) {
				int j = e[i];
				d[j]--;
				if(d[j]==0) {
					q.add(j);
				}
			}
		}
		return n==cnt;
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		n = sc.nextInt();
		m = sc.nextInt();
		for(int i=1;i<=m;i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int w = sc.nextInt();
			add(a,b,w);
            d[b] += 1;
		}
	}