poj 3013 Big Christmas Tree 最短路 dijkstra算法

题意：每个顶点有一个权值，求构造一个图，使得 ∑(每个点的权值)X(到1结点的距离) 最小

方法：dijkstra+heap

昨晚写了一个，不知道为什么TLE，怎么改都TLE。今天重写一遍AC了

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

#define MAXN 50005
#define MAXM 100005
#define INF 10000000000
int edge_cnt;
int val[MAXN];

int first[MAXN];
struct edge_node
{
	int to,weight,next;
}edges[MAXM];

inline void addedge(int f,int t,int dis)
{
	++edge_cnt;
	edges[edge_cnt].to=f;
	edges[edge_cnt].weight=dis;
	edges[edge_cnt].next=first[t];
	first[t]=edge_cnt;
	++edge_cnt;
	edges[edge_cnt].to=t;
	edges[edge_cnt].weight=dis;
	edges[edge_cnt].next=first[f];
	first[f]=edge_cnt;
}

struct Node
{
	int point,dist;
	bool operator<(const Node x) const
	{
		return x.dist<dist;
	}
};

bool done[MAXN];
long long dis[MAXN];
int n,m;
inline void dijkstra()
{
	memset(done,0,sizeof(done));
	priority_queue<Node> Que;
	Node tmp,inq;
	tmp.point=1;tmp.dist=0;
	for(int i=0;i<=n;i++)
		dis[i]=INF;
	dis[1]=0;
	Que.push(tmp);
	int u,v,e;
	while(!Que.empty())
	{
		tmp=Que.top();Que.pop();
		u=tmp.point;
		if(done[u])
			continue;
		done[u]=true;
		for(e=first[u];e;e=edges[e].next)
		{
			v=edges[e].to;
			if(!done[v]&&dis[v]>dis[u]+edges[e].weight)
			{
				dis[v]=dis[u]+edges[e].weight;
				inq.dist=dis[v];
				inq.point=v;
				Que.push(inq);
			}
		}
	}
}

int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		edge_cnt=0;

		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
			scanf("%d",&val[i]);
		int tmpf,tmpt,tmpd;
		memset(first,0,sizeof(first));
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d",&tmpf,&tmpt,&tmpd);
			addedge(tmpf,tmpt,tmpd);
		}
		dijkstra();
		long long ans=0;
		for(int i=1;i<=n;i++)
		{
			if(dis[i]==INF)
			{
				ans=-1;
				break;
			}
			ans+=dis[i]*val[i];
		}
		if(ans==-1)
			printf("No Answer\n");
		else
			printf("%lld\n",ans);
	}
	return 0;
}