Poj3013 big Christmas tree --- Shortest Path

I'm sorry to write it on the title. This is the shortest path.

This question is quite interesting. The key is to convert the question quantity into the shortest path.

Question:

For an undirected graph, each node has the right value p [I], and each edge has the right value W [I].

The minimum cost for connecting all vertices of the tree to the root node 1,

Minimum Cost = Σ W [I] × Σ p [I]

The first Σ is all edges, and the second Σ is the weight and

Ideas:

Through derivation, we can find that for each node, the cost of calculation is P [I] * d [I],

D [I] indicates the distance from the node to the root node.

So it suddenly becomes clear .. Question 1

The only pitfall is int64, and INF is large enough.

# Include <cstdio> # include <cstring> # include <queue> # include <iostream> using namespace STD; const ll INF = 1ll <60; // pitfall! This question INF must be large enough # define n 50010 # define M 100010 struct node {int V, next; ll W;} e [m]; int vis [N], n, m, head [N], H; ll d [N], p [N]; void addedge (int A, int B, ll c) {e [H]. V = B; E [H]. W = C; E [H]. next = head [a]; head [a] = H ++;} int spfa (INT st) {memset (VIS, 0, sizeof vis ); for (INT I = 0; I <= N; I ++) d [I] = inf; d [st] = 0; vis [st] = 1; queue <int> q; q. push (ST); int X, I, V; while (! Q. Empty () {x = Q. Front (); q. Pop (); vis [x] = 0; for (I = head [X]; I! =-1; I = E [I]. next) {v = E [I]. v; If (d [v]> d [x] + E [I]. w) {d [v] = d [x] + E [I]. w; If (! Vis [v]) {vis [v] = 1; q. push (v) ;}}}} void Init () {memset (Head,-1, sizeof head); H = 0 ;}int main () {int t, a, B, I; ll ans, C; scanf ("% d", & T); While (t --) {Init (); scanf ("% d", & N, & M); for (I = 1; I <= N; I ++) CIN> P [I]; while (M --) {scanf ("% d", & A, & B); CIN> C; addedge (A, B, C); addedge (B, a, c) ;}spfa (1); ans = 0; int flag = 1; for (I = 2; I <= N; I ++) {If (d [I] = inf) {flag = 0; break;} ans + = (d [I] * P [I]);} If (! Flag) printf ("No answer \ n"); else cout <ans <Endl;} return 0 ;}