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Dijkstra算法找单源最短路径

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//Dijkstra寻找最短路径（广度优先思想）
//用邻接矩阵实现图的存储，为简化问题，省去对图的定义
//单源最短路径问题：给定源顶点s，求它到其他任意顶点（目的顶点）的最短路径

#include<iostream>
using namespace std;
int Max = 100;
void djstr(int** a, int v, int e)
{
	int* p = new int[v];
	//p[i]：最短路径中顶点i的前驱顶点,
	//从终点开始，逆向即可获取路径
	int* d = new int[v];
	//当前最短路径长度

	int s = 0;
	bool* label = new bool[v];
	for (int i = 0; i < v; i++)
		label[i] = false;

	for (int i = 0; i < v; i++)
		if (a[0][i] > 0)
		{
			p[i] = 0;
			d[i] = a[0][i];
		}		
		else
		{
			p[i] = -1;
			d[i] = Max;
		}
	label[0] = true;
	while (s < v)
	{
		int index=0, min=d[0];
		for(int i=0;i<v;i++)
			if (!label[i]&&d[i] < min)
			{
				min = d[i];
				index = i;
			}
		++s; label[index] = true;
		for (int i = 0; i < v; i++)
		{
			if (a[index][i] + d[index] < d[i])
			{
				d[i] = a[index][i] + d[index];
				p[i] = index;
			}
		}
	}
	d[0] = 0;
	for (int i = 0; i < v; i++)
		cout << d[i] << " ";
	cout << endl;
	for (int i = 0; i < v; i++)
		cout << p[i]+1 << " ";
}
int main()
{
	
	int v = 5;
	int e = 7;
	int b[5][5] = {
	{ 0,4,2,Max,8 },
	{ Max,0,Max,4,5 },
	{ Max,Max,0,1,Max },
	{ Max,Max,Max,0,3 },
	{ Max,Max,Max,Max,0 } };
	int** a = new int*[v];
	for (int i = 0; i < v; i++)
	{
		a[i] = new int[v];
		for (int j = 0; j < v; j++)
			a[i][j] = b[i][j];
	}
	djstr(a, v, e);
	system("pause");
	return 0;
}