图的广度优先遍历

#include<iostream>
#include<algorithm>
#include<cstdlib>
using namespace std; 
const int maxn = 9999999;
struct node {
	int x, s;
};

struct node que[2501];
int e[101][101], book[51];
int main() {
	int head, tail;
	int n, m, a,b, cur, start, end, flag = 0;
	cin >> n >> m >> start >> end;
	for(int i = 1; i<=n;i++)
		for (int j = 1; j <= n; j++)
		{
			if (i == j) e[i][j] = 0;
			else e[i][j] = maxn;
		}
	for (int i = 1; i <= m; i++) {
		cin >> a >> b;
		e[a][b] = 1;
		e[b][a] = 1;//无向图
	}
	head = 1;
	tail = 1;
	que[head].x = start;
	que[head].s = 0;
	tail++;
	book[start] = 1;
	while (head < tail) {
		cur = que[head].x;
		for (int j = 1; j <= n; j++) {
			if (e[cur][j] != maxn && book[j] == 0) {
				que[tail].x = j;
				que[tail].s = que[head].s + 1;
				tail++;
				book[j] = 1;
			}
			if (que[tail - 1].s == end) {
				flag = 1;
				break;
			}
		}
		if (flag == 1)
			break;
		head++;
	}
	printf("%d", que[tail - 1].s); 
	return 0;
}

​ 本题也可使用深度优先搜索，但是这里使用广度优先搜索会更快。广度优先搜索会更加适用于所有边的 权值相同的情况。