差分约束系统 Dijkstra模板（2）

/**差分约束 Dijkstra 模板（数组）这题用vector会超时，后来才知道这个叫链式前向星差分约束感觉这个博客写的比较好：http://blog.csdn.net/xuezhongfenfei/article/details/8685313**/#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>using namespace std;const int INF=1<<27;const int maxn=30005;struct HeapNode//优先队列的节点{   int d,u;   bool operator < (const HeapNode& rhs) const   {      return d>rhs.d;   }};struct Edge{   int next,to,dist; //next表示从同一个点出发相邻的边}edges[150005];struct Dijkstra{   int n,m,total;//点数和边数   int head[maxn]; //每个点出发的边编号（从0开始编号）   bool done[maxn]; //是否已经永久标号   int d[maxn]; //s到各个点之间的距离   int p[maxn]; //最短路中的上一条边   void init(int n)   {      this->n=n;      memset(head,-1,sizeof(head));      total=0;   }   void AddEdge(int from,int to,int dist)   {      edges[total].to=to;      edges[total].dist=dist;      edges[total].next=head[from];      head[from]=total++;   }   void dijkstra(int s)//求s到所有点的距离   {      priority_queue<HeapNode> Q;      for(int i=0;i<n;i++) d[i]=INF;      d[s]=0;      memset(done,0,sizeof(done));      Q.push((HeapNode){0,s});      while(!Q.empty())      {         HeapNode x=Q.top();         Q.pop();         int u=x.u;         if(done[u]) continue;         done[u]=true;         for(int i=head[u];i!=-1;i=edges[i].next)         {            Edge& e=edges[i];            if(d[e.to]>d[u]+e.dist)            {               d[e.to]=d[u]+e.dist;               p[e.to]=u;               Q.push((HeapNode){d[e.to],e.to});            }         }      }   }};int main(){   int n,m,i,x,y,z;   Dijkstra solver;   scanf("%d%d",&n,&m);   solver.init(n);   for(i=0;i<m;i++)   {      scanf("%d%d%d",&x,&y,&z);      solver.AddEdge(x-1,y-1,z);   }   solver.dijkstra(0);   printf("%d\n",solver.d[n-1]);   return 0;}