HDU 4725 The Shortest Path in Nya Graph (最短路)
The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 37 Accepted Submission(s): 6
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
Recommend
zhuyuanchen520
最短路。
主要是建图。
N个点,然后有N层,要假如2*N个点。
总共是3*N个点。
点1~N就是对应的实际的点1~N. 要求的就是1到N的最短路。
然后点N+1 ~ 3*N 是N层拆出出来的点。
第i层,入边到N+2*i-1, 出边从N+2*i 出来。(1<= i <= N)
N + 2*i 到 N + 2*(i+1)-1 加边长度为C. 表示从第i层到第j层。
N + 2*(i+1) 到 N + 2*i - 1 加边长度为C,表示第i+1层到第j层。
如果点i属于第u层,那么加边 i -> N + 2*u -1 N + 2*u ->i 长度都为0
然后用优先队列优化的Dijkstra就可以搞出最短路了
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-9-11 12:30:12 4 File Name :2013-9-11\1010.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 /* 22 * 使用优先队列优化Dijkstra算法 23 * 复杂度O(ElogE) 24 * 注意对vector<Edge>E[MAXN]进行初始化后加边 25 */ 26 const int INF=0x3f3f3f3f; 27 const int MAXN=1000010; 28 struct qnode 29 { 30 int v; 31 int c; 32 qnode(int _v=0,int _c=0):v(_v),c(_c){} 33 bool operator <(const qnode &r)const 34 { 35 return c>r.c; 36 } 37 }; 38 struct Edge 39 { 40 int v,cost; 41 Edge(int _v=0,int _cost=0):v(_v),cost(_cost){} 42 }; 43 vector<Edge>E[MAXN]; 44 bool vis[MAXN]; 45 int dist[MAXN]; 46 void Dijkstra(int n,int start)//点的编号从1开始 47 { 48 memset(vis,false,sizeof(vis)); 49 for(int i=1;i<=n;i++)dist[i]=INF; 50 priority_queue<qnode>que; 51 while(!que.empty())que.pop(); 52 dist[start]=0; 53 que.push(qnode(start,0)); 54 qnode tmp; 55 while(!que.empty()) 56 { 57 tmp=que.top(); 58 que.pop(); 59 int u=tmp.v; 60 if(vis[u])continue; 61 vis[u]=true; 62 for(int i=0;i<E[u].size();i++) 63 { 64 int v=E[tmp.v][i].v; 65 int cost=E[u][i].cost; 66 if(!vis[v]&&dist[v]>dist[u]+cost) 67 { 68 dist[v]=dist[u]+cost; 69 que.push(qnode(v,dist[v])); 70 } 71 } 72 } 73 } 74 void addedge(int u,int v,int w) 75 { 76 E[u].push_back(Edge(v,w)); 77 } 78 79 int main() 80 { 81 //freopen("in.txt","r",stdin); 82 //freopen("out.txt","w",stdout); 83 int T; 84 int N,M,C; 85 scanf("%d",&T); 86 int iCase = 0; 87 while(T--) 88 { 89 scanf("%d%d%d",&N,&M,&C); 90 for(int i = 1;i <= 3*N;i++) E[i].clear(); 91 int u,v,w; 92 for(int i = 1;i <= N;i++) 93 { 94 scanf("%d",&u); 95 addedge(i,N + 2*u - 1,0); 96 addedge(N + 2*u ,i,0); 97 98 } 99 for(int i = 1;i < N;i++) 100 { 101 addedge(N + 2*i-1,N + 2*(i+1),C); 102 addedge(N + 2*(i+1)-1,N + 2*i,C); 103 } 104 while(M--) 105 { 106 scanf("%d%d%d",&u,&v,&w); 107 addedge(u,v,w); 108 addedge(v,u,w); 109 } 110 Dijkstra(3*N,1); 111 iCase++; 112 if(dist[N] == INF)dist[N] = -1; 113 printf("Case #%d: %d\n",iCase,dist[N]); 114 115 } 116 return 0; 117 }
人一我百!人十我万!永不放弃~~~怀着自信的心,去追逐梦想
posted on 2013-09-11 17:44 kuangbin 阅读(4017) 评论(11) 编辑 收藏 举报