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.
 

 

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.
 

 

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.
 

 

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
2013 ACM/ICPC Asia Regional Online —— Warmup2
 

 

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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。
  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 }
View Code

 

posted @ 2014-05-08 02:41  hyx1  阅读(383)  评论(0)  编辑  收藏  举报