POJ 3013 Dijkstra

从1节点最短路，，然后再乘一下权值就OK了

//By SiriusRen#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 132000long long d[N];int n,m,first[N],next[N],v[N],w[N],tot,a[N],xx,yy,zz,cases;bool vis[N];struct Node{int now,weight;}jy;void add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}bool operator < (Node a,Node b){return a.weight>b.weight;}void Dijkstra(){    priority_queue<Node>pq;    memset(d,0x3f,sizeof(d)),memset(vis,0,sizeof(vis)),d[1]=0;    jy.now=1,jy.weight=0;pq.push(jy);    while(!pq.empty()){        Node t=pq.top();pq.pop();        if(!vis[t.now])vis[t.now]=1;        else continue;        for(int i=first[t.now];~i;i=next[i])            if(!vis[v[i]]&&d[v[i]]>d[t.now]+w[i]){                d[v[i]]=d[t.now]+w[i];                jy.now=v[i],jy.weight=d[v[i]];                pq.push(jy);            }    }    long long ans=0;    for(int i=1;i<=n;i++)        if(d[i]>0x3ffffffff){puts("No Answer");return;}        else ans+=d[i]*a[i];    printf("%lld\n",ans);}signed main(){    scanf("%d",&cases);    while(cases--){        tot=0;memset(first,-1,sizeof(first));        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        for(int i=1;i<=m;i++){            scanf("%d%d%d",&xx,&yy,&zz);            add(yy,xx,zz),add(xx,yy,zz);        }        Dijkstra();    }}