圖論演算法(一)--最短路徑的DFS/BFS解法(JAVA )
阿新 • • 發佈:2019-01-05
最短路徑–城市路徑問題:
問題描述:求從1號城市到5號城市的最短路徑長度
Input:
5 8
1 2 2
1 5 10
2 3 3
2 5 7
3 1 4
3 4 4
4 5 5
5 3 3
Output:
9
DFS
import java.util.Scanner;
public class minPath {
static int min = 99999999;
static int[][] e = new int[100][100];
static int[] book = new int[100];
static int n, m;
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();
int c = input.nextInt();
e[a][b] = c;
}
book[1 ] = 1;
dfs(1, 0);
System.out.println(min);
}
public static void dfs(int cur, int dis) {
/**
* 如果當前路徑大於之前找到的最小值,可直接返回
* */
if (dis > min) {
return;
}
/**
* 判斷是否達到最後一個結點,更新最小值,返回
* */
if(cur == n) {
if (dis < min) {
min = dis;
return;
}
}
/**
* 當前點到其他各點之間可連通但是還未新增進來時,遍歷執行
* */
for (int i = 1; i <= n; i++) {
if (e[cur][i] != 99999999 && book[i] == 0) {
book[i] = 1;
dfs(i, dis+e[cur][i]);
/**
* 回溯
**/
book[i] = 0;
}
}
return;
}
}
最短路徑–轉乘問題:
問題描述:求從1號城市到5號城市的最短路徑長度
Input:
5 7 1 5
1 2
1 3
2 3
2 4
3 4
3 5
4 5
Output:
2
這個題目與上一個的區別就在於,這些路徑是沒有權值的,也就是說只需要找出到達一個點的最短距離就可以了,記錄次數即可。
BFS
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
class node {
int x;
int s;
node(int x, int s) {
this.x = x;
this.s = s;
}
}
public class minPath {
static int[][] e = new int[51][51];
static int[] book = new int[51];
static int n, m;
static int start, end;
static int mark, sum;
static Queue<node> queue = new LinkedList<>();
static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
n = input.nextInt();
m = input.nextInt();
start = input.nextInt();
end = input.nextInt();
for (int i = 1; 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++) {
int a = input.nextInt();
int b = input.nextInt();
e[a][b] = 1;
e[b][a] = 1;
}
queue.offer(new node(start, 0));
book[1] = start;
bfs();
System.out.println(sum);
}
public static void bfs() {
int flag = 0;
while (!queue.isEmpty()) {
int cur = queue.peek().x;
for (int i = 1; i <= n; i++) {
if(e[cur][i] != 99999999 && book[i] == 0) {
mark = i;
sum = queue.peek().s + 1;
queue.offer(new node(i, sum));
book[i] = 1;
}
if(mark == end) {
flag = 1;
break;
}
}
if(flag == 1) {
break;
}
queue.remove();
}
return;
}
}
基本上能用深度優先的問題都可以用廣度優先,但是廣度優先更適合無向圖,或者說所有邊的權值一樣的情況,大家可以通過多做題靈活使用這兩種方法