搜素演算法(基礎)--DFS/BFS演算法(JAVA)
阿新 • • 發佈:2018-12-30
為了便於理解這裡的資料是一個無向圖,要求輸出遍歷順序
下面只給出用例和演算法,之後可以根據後面的三個題目進行深入學習
Input:
5 5
1 2
1 3
1 5
2 4
3 5
Output:
1 2 4 3 5
DFS
import java.util.Scanner;
public class DFS {
static int[][] e = new int[100][100];
static int[] book = new int[100];
static int n, m;
static int sum = 0;
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
n = input.nextInt();
m = input.nextInt();
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (i == j) {
e[i][j] = 0;
} else {
e[i][j] = 99999999 ;
}
}
}
for (int i = 1; i <= m; i++) {
int a = input.nextInt();
int b = input.nextInt();
e[a][b] = 1;
e[b][a] = 1;
}
book[1] = 1;
dfs(1);
}
public static void dfs(int cur) {
System.out .print(cur + " ");
sum++;
if (sum == n) {
return;
}
for (int i = 1; i <= n; i++) {
if (e[cur][i] == 1 && book[i] == 0) {
book[i] = 1;
dfs(i);
}
}
return;
}
}
BFS
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class BFS {
static int[][] e = new int[100][100];
static int[] book = new int[100];
static int n, m;
static Queue<Integer> queue = new LinkedList<>();
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
int a, b;
n = input.nextInt();
m = input.nextInt();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == j) {
e[i][j] = 0;
} else {
e[i][j] = 99999999;
}
}
}
for (int i = 1; i <= m; i++) {
a = input.nextInt();
b = input.nextInt();
e[a][b] = 1;
e[b][a] = 1;
}
queue.offer(1);
book[1] = 1;
bfs();
}
public static void bfs() {
while (!queue.isEmpty()) {
int cur = queue.peek();
for (int i = 1; i <= n; i++) {
if (e[cur][i] == 1 && book[i] == 0) {
book[i] = 1;
queue.offer(i);
}
}
System.out.print(queue.remove() + " ");
}
return;
}
}
下面是一個全排列的簡單題目,也用到了dfs,屬於偏簡單的一道題目,用來給後面的題目做鋪墊
問題描述:假設有編號為1、2、3的三張卡片和編號1、2、3的三個盒子,現需將這三張撲克牌分別放到三個盒子裡面,並且每個盒子只能放一張撲克牌,請問一共有幾種不同的方法?
Output:
123 132 213 231 312 321
import java.util.Scanner;
/**
* DFS演算法基本模型:
* void dfs(int step) {
* 判斷邊界
* 嘗試每一種可能 for {
* 繼續下一步def(step + 1)
* }
* 返回
* }
* */
public class DFS {
static int n;
static int[] book = new int[10];
static int[] a = new int[10];
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
n = input.nextInt();
dfs(1);
}
public static void dfs(int step) {
if (step == n + 1) {
/**
* 列印,並返回
* */
for (int i = 1; i <= n; i++) {
System.out.print(a[i] + " ");
}
System.out.println();
return;
}
for (int i = 0; i < n; i++) {
/**
* 標記陣列book,用來檢測盒子中是否已經放入
* */
if (book[i] == 0) {
/**
* 將i號撲克牌放入到第step個盒子中
* */
a[step] = i;
book[i] = 1;
/**
* 處理step+1個小盒子,注意要記得將剛才嘗試的撲克牌收回
* */
dfs(step + 1);
book[i] = 0;
}
}
return;
}
}