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;

}