迪杰斯特拉的优化 复杂度e*logv

const int MAXN = 1e2 + 5;
int a[N][N];
int dis[N];
int vis[N];
int n, m, s, t;
int x, y, v;
struct edge {
	int to;//终点
	int cost;//花费
	edge(int x, int y) :to(x),cost(y) {}
};
typedef pair<int, int> P;//first为最短距离，second为定点编号。
vector<edge> graph[MAXN];
int dis[MAXN];
int path[MAXN];
int dij(int s)
{
	memset(path, -1, sizeof(path));
	memset(dis, 0x3f, sizeof(dis));
	dis[s] = 0;
	priority_queue<P, vector<P>, greater<P>> que;
	que.push(P(0, s));
	while (!que.empty())
	{
		P p = que.top();
		que.pop();
		int v = p.second;
		if (dis[v] < p.first) continue;
		for (int i = 0; i < graph[v].size(); i++)
		{
			edge e = graph[v][i];
			if (dis[e.to] > dis[v] + e.cost)
			{
				dis[e.to] = dis[v] + e.cost;
				que.push(P{ dis[e.to],e.to });
				path[e.to] = v;
			}
		}
	}
}