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
挺有意思的最短路 题意是在无向图中，每个节点还属于一个层（layer），可以从第i个层的节点移动到第i+1个层的节点，花费是c。然后是给出一个普通的无向图，求1到n的最短路。 因为有层的存在，在寻找最短路的时候可能直接经过层，不妨把层也当做节点，层与所属节点之间的花费是0，表示随时可以从层转移到节点上寻找最短路，然后层与层之间的花费是c，节点到下一层或上一层的的花费也是c，节点与节点之间的花费是w，方便起见，层的节点编号从n+1到n+n。 另外数组开大点，不然会TLE，我很奇怪为什么不是RE，害得我判断失误
#include <iostream> #include <cstring> #include <string> #include <vector> #include <queue> #include <cstdio> #include <set> #include <cmath> #include <algorithm> #define INF 0x3f3f3f3f #define MAXN 200010 #define Mod 10001 using namespace std; struct Edge { int v,w,next; }; Edge edge[MAXN*20]; int head[MAXN],n,m,e,vis[MAXN],dis[MAXN]; void add(Edge *edge,int *head,int u,int v,int w) { edge[e].v=v; edge[e].w=w; edge[e].next=head[u]; head[u]=e; e++; } void spfa(Edge *edge,int *head,int u) { memset(vis,0,sizeof(vis)); for(int i=1; i<=2*n; ++i) //这里WA了 dis[i]=INF; dis[u]=0; queue<int> q; q.push(u); while(!q.empty()) { u=q.front(); q.pop(); vis[u]=0; for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].v,w=edge[i].w; if(w+dis[u]<dis[v]) { dis[v]=w+dis[u]; if(!vis[v]) { vis[v]=1; q.push(v); } } } } } int layer[MAXN]; int main() { int t,c,u,v,w; scanf("%d",&t); for(int cnt=1; cnt<=t; ++cnt) { e=0; scanf("%d%d%d",&n,&m,&c); memset(vis,0,sizeof(vis)); memset(head,-1,sizeof(head)); for(int i=1; i<=n; ++i) { scanf("%d",&u); layer[i]=u; vis[u]=1; } for(int i=1; i<n; ++i) { if(vis[i]&&vis[i+1]) { add(edge,head,i+n,i+n+1,c); add(edge,head,i+n+1,i+n,c); } } for(int i=1;i<=n;++i) { add(edge,head,n+layer[i],i,0); //add(edge,head,i,n+layer[i],0); 有多个节点对应一个层的情况，大概。。。 if(layer[i]>1) add(edge,head,i,n+layer[i]-1,c); if(layer[i]<n) add(edge,head,i,n+layer[i]+1,c); } for(int i=1;i<=m;++i) { scanf("%d%d%d",&u,&v,&w); add(edge,head,u,v,w); add(edge,head,v,u,w); } spfa(edge,head,1); if(dis[n]<INF) printf("Case #%d: %d\n",cnt,dis[n]); else printf("Case #%d: -1\n",cnt); } return 0; }