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;
}

  

posted @   laputastar  阅读(245)  评论(0编辑  收藏  举报
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