hdu 4725
The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2296 Accepted Submission(s): 561
Problem Description
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.
Input
The 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 <= 105) and C(1 <= C <= 103), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers li (1 <= li <= N), which is the layer of ith node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 104), which means there is an extra edge, connecting a pair of node u and v, with cost w.
For each test case, first line has three numbers N, M (0 <= N, M <= 105) and C(1 <= C <= 103), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers li (1 <= li <= N), which is the layer of ith node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 104), which means there is an extra edge, connecting a pair of node u and v, with cost w.
Output
For 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.
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
Source
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最短路
把每一层拆成2 个点 一个是连接结点入边,一个是连接结点出边,每层连接结点入边的点连接下个层连接结点出边的点,下一层连接入边的点连接上一层连接出边的点。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 8 using namespace std; 9 10 const int MAX_N = 3e5 + 7; 11 const int INF = 1e9 + 7; 12 typedef long long ll; 13 struct heapnode { 14 int d; 15 int u; 16 bool operator < (const heapnode &rhs) const { 17 return d > rhs.d; 18 } 19 }; 20 struct Edge {int from, to, cost;}; 21 22 vector<int> G[MAX_N]; 23 vector<Edge> edges; 24 int N,M,C; 25 int d[MAX_N]; 26 bool done[MAX_N]; 27 vector <int> lay[MAX_N]; 28 29 void add_edge(int from, int to, int cost) { 30 edges.push_back((Edge){from, to, cost}); 31 int m = edges.size(); 32 G[from].push_back(m - 1); 33 } 34 35 void dij(int s) { 36 memset(done,0,sizeof(done)); 37 for(int i = 1; i <= 3 * N; ++i) { 38 d[i] = INF; 39 } 40 d[s] = 0; 41 priority_queue <heapnode> q; 42 q.push((heapnode) {d[s], s}); 43 44 while(!q.empty()) { 45 heapnode x = q.top(); q.pop(); 46 int u = x.u; 47 if(done[u]) continue; 48 done[u] = 1; 49 if(u == N) return; 50 51 for(int i = 0; i < G[u].size(); ++i) { 52 Edge &e = edges[ G[u][i] ]; 53 if(d[e.to] > d[u] + e.cost) { 54 d[e.to] = d[u] + e.cost; 55 q.push((heapnode) {d[e.to], e.to}); 56 } 57 } 58 } 59 } 60 int main() 61 { 62 63 // freopen("sw.in","r",stdin); 64 int t; 65 scanf("%d",&t); 66 int ca = 1; 67 while(t--) { 68 scanf("%d%d%d",&N,&M,&C); 69 for(int i = 1; i <= 3 * N; ++i) G[i].clear(); 70 edges.clear(); 71 72 for(int i = 1; i <= N; ++i) { 73 int ch; 74 scanf("%d",&ch); 75 add_edge(i, ch + N, 0); 76 add_edge(ch + 2 * N, i, 0); 77 78 } 79 80 /*for(int i = 2 * N + 1; i <= 3 * N; ++i) { 81 add_edge(i, i - N, 0); 82 }*/ 83 84 85 for(int i = N + 1; i <= 2 * N - 1; ++i) { 86 add_edge(i, i + 1 + N, C); 87 add_edge(i + 1, i + N, C); 88 } 89 90 for(int i = 1; i <= M; ++i) { 91 int u, v, w; 92 scanf("%d%d%d",&u, &v, &w); 93 add_edge(u, v, w); 94 add_edge(v, u, w); 95 } 96 97 dij(1); 98 printf("Case #%d: %d\n",ca++, (d[N] == INF) ? -1 : d[N]); 99 } 100 101 return 0; 102 }