POJ 3013 Big Christmas Tree

Big Christmas Tree
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 23525 Accepted: 5104
Description

Christmas is coming to KCM city. Suby the loyal civilian in KCM city is preparing a big neat Christmas tree. The simple structure of the tree is shown in right picture.

The tree can be represented as a collection of numbered nodes and some edges. The nodes are numbered 1 through n. The root is always numbered 1. Every node in the tree has its weight. The weights can be different from each other. Also the shape of every available edge between two nodes is different, so the unit price of each edge is different. Because of a technical difficulty, price of an edge will be (sum of weights of all descendant nodes) × (unit price of the edge).

Suby wants to minimize the cost of whole tree among all possible choices. Also he wants to use all nodes because he wants a large tree. So he decided to ask you for helping solve this task by find the minimum cost.

Input

The input consists of T test cases. The number of test cases T is given in the first line of the input file. Each test case consists of several lines. Two numbers v, e (0 ≤ v, e ≤ 50000) are given in the first line of each test case. On the next line, v positive integers wi indicating the weights of v nodes are given in one line. On the following e lines, each line contain three positive integers a, b, c indicating the edge which is able to connect two nodes a and b, and unit price c.

All numbers in input are less than 216.

Output

For each test case, output an integer indicating the minimum possible cost for the tree in one line. If there is no way to build a Christmas tree, print “No Answer” in one line.

Sample Input

2
2 1
1 1
1 2 15
7 7
200 10 20 30 40 50 60
1 2 1
2 3 3
2 4 2
3 5 4
3 7 2
3 6 3
1 5 9
Sample Output

15
1210
题意：给你一个计算权值的公式，然后计算一条最短路的权值
直接建立邻接表，正反处理点，然后直接模板spfa，统计权值就行了

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <queue>const int VM=50010;const int EM=50010;const long long INF=9999999999;#define LL long longusing namespace std;int n,m,k,wight[VM];int head[VM],vist[VM];long long d[VM];struct node{    int v,next;    long long w;}map[EM<<1];void addmap(int u,int v,long long w){    map[k].v=v;    map[k].w=w;    map[k].next=head[u];    head[u]=k++;}LL spfa(){    int i;    queue<int>q;    while(!q.empty())        q.pop();    for(i=1; i<=n; i++)    {        d[i]=INF;        vist[i]=0;    }    d[1]=0;    vist[1]=1;    q.push(1);    while(!q.empty())    {        int now=q.front();        q.pop();        vist[now]=0;        for(i=head[now]; i!=-1; i=map[i].next)        {            int v=map[i].v;            if(d[v]>d[now]+map[i].w)            {                d[v]=d[now]+map[i].w;                if(!vist[v])                {                    vist[v]=1;                    q.push(v);                }            }        }    }    LL ans=0;    for(i=1; i<=n; i++)    {        if(d[i]==INF)            return 0;        ans+=wight[i]*d[i];    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        k=0;        memset(head,-1,sizeof(head));        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)            scanf("%d",&wight[i]);        int u,v;        LL w;        while(m--)        {            scanf("%d%d%I64d",&u,&v,&w);            addmap(u,v,w);            addmap(v,u,w);        }        if(n==0 || n==1)        {            printf("0\n");            continue;        }        long long ans=spfa();        if(ans==0)            printf("No Answer\n");        else            printf("%I64d\n",ans);    }    return 0;}