The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13070 Accepted Submission(s): 2794
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4725
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
题意：
给出一个分层图，每个点只属于一层，点与点之间到达有一定花费，然后相邻两层移动也有一定花费。最后问从1号点到n号点的最小花费是什么。
题解：
朴素的想法就是点与点之间连边，层与层之间连边，但是空间会爆掉。
于是就像将“层”给分离出去，如果点i属于x层，就往n+x连边，这样会节约很多的空间。
但是这也有一个问题，就是跑最短路的时候点跑向相应的层，然后又跑回来。
所以我们考虑只连出边/入边，然后层直接与点相连。
具体代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 2e5+;
int n,m,c,t,tot;
int head[N],d[N],vis[N],belong[N],lay[N];
struct Edge{
int v,w,next ;
}e[N<<];
struct node{
int d,u;
bool operator < (const node &A)const{
return d>A.d;
}
};
void adde(int u,int v,int w){
e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
}
void Dijkstra(int s){
priority_queue <node> q;memset(d,INF,sizeof(d));
memset(vis,,sizeof(vis));d[s]=;
q.push(node{,s});
while(!q.empty()){
node cur = q.top();q.pop();
int u=cur.u;
if(vis[u]) continue ;
vis[u]=;
for(int i=head[u];i!=-;i=e[i].next){
int v=e[i].v;
if(d[v]>d[u]+e[i].w){
d[v]=d[u]+e[i].w;
q.push(node{d[v],v});
}
}
}
}
int main(){
cin>>t;
int cnt =;
while(t--){
cnt++;
scanf("%d%d%d",&n,&m,&c);
memset(head,-,sizeof(head));tot=;
memset(belong,,sizeof(belong));
memset(lay,,sizeof(lay));
for(int i=,tmp;i<=n;i++){
scanf("%d",&tmp);
lay[tmp]=;
belong[i]=tmp;
}
for(int i=;i<=n;i++){
if(!lay[i] || !lay[i-]) continue ;
adde(i+n,i+n-,c);
adde(i+n-,i+n,c);
}
for(int i=;i<=n;i++){
int tmp = belong[i];
adde(n+tmp,i,);
if(tmp>) adde(i,tmp+n-,c);
if(tmp<n) adde(i,tmp+n+,c);
//adde(i,n+tmp,0);
}
for(int i=,u,v,w;i<=m;i++){
scanf("%d%d%d",&u,&v,&w);
adde(u,v,w);adde(v,u,w);
}
Dijkstra();
printf("Case #%d: ",cnt);
if(d[n]==INF) puts("-1");
else cout<<d[n]<<endl;
}
return ;
}