题目：
This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on.
The Nya graph is an undirected graph with "layers". Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.
The Nya graph is an undirected graph with "layers". Each node in the graph belongs to a layer, there are N nodes in total.
You can move from any node in layer x to any node in layer x + 1, with cost C, since the roads are bi-directional, moving from layer x + 1 to layer x is also allowed with the same cost.
Besides, there are M extra edges, each connecting a pair of node u and v, with cost w.
Help us calculate the shortest path from node 1 to node N.
InputThe first line has a number T (T <= 20) , indicating the number of test cases.
For each test case, first line has three numbers N, M (0 <= N, M <= 10 5) and C(1 <= C <= 10 3), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l i (1 <= l i <= N), which is the layer of i th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 4), which means there is an extra edge, connecting a pair of node u and v, with cost w.OutputFor test case X, output "Case #X: " first, then output the minimum cost moving from node 1 to node N.
If there are no solutions, output -1.Sample Input
2
3 3 3
1 3 2
1 2 1
2 3 1
1 3 3
3 3 3
1 3 2
1 2 2
2 3 2
1 3 4
Sample Output
Case #1: 2
Case #2: 3
分析：
主要是构图，m条边正常建，对于相邻层的建图：对于每一层，层点 → 该层每个点，该层每个点→邻层层点（均为单项）
代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 9e5 + 10; 5 const ll inf = 1e18; 6 struct edge 7 { 8 int e, v, next; 9 edge(int a = 0, int b = 0, int c = 0) : e(a), v(b), next(c) {} 10 }e[maxn]; 11 int head[maxn]; 12 ll dis[maxn]; 13 int vis[maxn]; 14 int v[maxn]; 15 int have[maxn]; 16 int tot; 17 18 struct node 19 { 20 int e; 21 ll dis; 22 node(int a = 0, ll b = 0) : e(a), dis(b) {} 23 bool operator< (const node& a) const 24 { 25 return a.dis < dis; 26 } 27 }; 28 29 void add(int x, int y, int v, int tot) 30 { 31 e[tot] = edge(y, v, head[x]); 32 head[x] =tot; 33 } 34 35 void dj() 36 { 37 for (int i = 1; i < maxn; i++) 38 dis[i] = inf; 39 dis[1] = 0; 40 memset(vis, 0, sizeof(vis)); 41 priority_queue<node> q; 42 q.push(node(1, 0)); 43 44 45 while (!q.empty()) 46 { 47 node temp = q.top(); q.pop(); 48 int u = temp.e; 49 if (vis[u]) continue; 50 vis[u] = 1; 51 for (int i = head[u]; i != -1; i = e[i].next) 52 { 53 int y = e[i].e; 54 if (!vis[y] && dis[y] > dis[u] + e[i].v) 55 { 56 dis[y] = dis[u] + e[i].v; 57 q.push(node(y, dis[y])); 58 } 59 } 60 } 61 } 62 63 int main() 64 { 65 int T; cin >> T; 66 int n, m, c; 67 int cases = 0; 68 69 while (T--) 70 { 71 int x, y, w; 72 73 cin >> n >> m >> c; 74 tot = 0; 75 memset(head, -1, sizeof(head)); 76 memset(have, 0, sizeof(have)); 77 for (int i = 1; i <= n; i++)//input 78 { 79 scanf("%d", &v[i]); 80 have[v[i]] = 1; 81 } 82 for (int i = 1; i <= n; i++) 83 { 84 add(v[i] + n, i, 0, ++tot); 85 if (v[i] > 1 && have[v[i] - 1]) add(i, v[i] - 1 + n, c, ++tot); 86 if (v[i] < n && have[v[i] + 1]) add(i, v[i] + 1 + n, c, ++tot); 87 } 88 for (int i = 1; i <= m; i++)//point to point 89 { 90 scanf("%d%d%d", &x, &y, &w); 91 add(x, y, w, ++tot); 92 add(y, x, w, ++tot); 93 } 94 dj(); 95 printf("Case #%d: %lld ", ++cases, dis[n] == inf ? -1 : dis[n]); 96 } 97 }