图解Dijkstra算法（单源最短路径）

注意这里这段话的理解，我们选择了距离源点最近的点之后，必然不可能再找到比它离源点还近的点了，例如我们找到距离源点最近的点距离是1，那么我们要想通过“中间点”找到比它还小的距离是不可能的，因为单凭这个1就是最小的了，何况我们还要加上另外一段距离

 

 

#include
#include
using namespace std;

int main()
{
	int e[10][10], dis[10], book[10], i, j, n, m, t1, t2, t3, u, v, minn;
	int inf = 99999999;
	cin >> n >> m;
	for ( i = 1; i <= n; i++)//初始化
	{
		for ( j = 1; j <= n; j++)
		{
			if (i == j)e[i][j] = 0;
			else e[i][j] = inf;
		}
	}
	for ( i = 1; i <= m; i++)//输入边
	{
		cin >> t1 >> t2 >> t3;
		e[t1][t2] = t3;
	}
	for ( i = 1; i <= n; i++)
	{
		dis[i] = e[1][i];
	}
	for ( i = 1; i <= n; i++)//初始化
	{
		book[i] = 0;
	}
	book[1] = 1;

	for ( i = 1; i <= n; i++)//核心代码
	{
		minn = inf;
		for ( j = 1; j <= n; j++)
		{
			if (book[j] == 0 && dis[j] < minn)
			{
				minn = dis[j];
				u = j;
			}
		}
		book[u] = 1;
		for (v = 1; v <= n; v++)
		{
			if (e[u][v] < inf)
			{
				dis[v] = min(dis[v], dis[u] + e[u][v]);
			}
		}
	}
	for ( i = 1; i <= n; i++)
	{
		cout << dis[i] << ' ';
	}
	return 0;
}

 以上内容出自啊哈算法