poj 3013 Big Christmas Tree 最短路 （转换思维，看点不看边）

Big Christmas Tree

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 24955 Accepted: 5393

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 w_i 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

22 11 11 2 157 7200 10 20 30 40 50 601 2 12 3 32 4 23 5 43 7 23 6 31 5 9

Sample Output

15
1210
如果想不通就不会做..想通了超级简单。
如图：A
    / \
   B   C
  /\   /\
 D E  F  G
按照题意我们分别算每一条边的值。
val（AB）=l（A，B）*（B+D+E）
val（BD）=l（B，D）*D
val（BE）=l（B,E）*E
val（AC）=l（A，C）*（C+F+G）
val（CF）=l（C,F）*F
val（CG）=l（C,G）*G
常规就是这样，按边算。
我们变成按点算，把上面公式按点提取。
val(B)=L(A,B)*B;
val(C)=L(A,C)*C
val(D)=(L(A,B)+L(B,D))*C
以此类推
我们会发现每个点对应的值就是A到这个点的最短路乘以这个点的权值。如果碰到INF，则不存在。
求即可
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
#define N 50005
#define INF (1LL<<60)


struct Edge
{
    int v,next;
    long long w;
} edge[N];


int head[N];


int cnte=0;
void addEdge(int a,int b,int w)
{
    edge[cnte].v=b;
    edge[cnte].w=w;
    edge[cnte].next=head[a];
    head[a]=cnte++;
}


long long dist[N];
bool vis[N];
bool Spfa(int n)
{
    for(int i=1; i<=n; i++)
    {
        dist[i]=INF;
        vis[i]=0;
    }
    dist[1]=0;
    vis[1]=1;
    queue<int>q;
    q.push(1);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(dist[v]>dist[u]+edge[i].w)
            {
                dist[v]=dist[u]+edge[i].w;
                if(vis[v]==0)
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
        vis[u]=0;
    }
    for(int i=1; i<=n; i++)
    {
        if(dist[i]==INF)
            return 0;
    }
    return 1;
}


int val[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        cnte=0;
        int vn,en;
        scanf("%d%d",&vn,&en);
        for(int i=1; i<=vn; i++)
            cin>>val[i];
        memset(head,-1,sizeof(head));
        for(int i=0; i<en; i++)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            addEdge(a,b,w);
            addEdge(b,a,w);
        }
        if(!Spfa(vn))
        {
            puts("No Answer");
            continue;
        }
        else
        {
            long long res=0;
            for(int i=1; i<=vn; i++)
                res+=val[i]*dist[i];
            printf("%lld\n",res);
        }
    }
    return 0;
}