The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13445 Accepted Submission(s): 2856

　

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 <= 10 ⁵) and C(1 <= C <= 10 ³), which is the number of nodes, the number of extra edges and cost of moving between adjacent layers.
The second line has N numbers l _i (1 <= l _i <= N), which is the layer of i ^th node belong to.
Then come N lines each with 3 numbers, u, v (1 <= u, v < =N, u <> v) and w (1 <= w <= 10 ⁴), 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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【题意】

给一张图，n个点，m条有权无向边，每个点属于某一层，相邻层间的任意两点存在一条权值为C的边，问1到n的最短路

【分析】

此图数据量比较大，暴力建图不可取！会MLE or TLE.

可以将层也抽象化成点，也就是一共有N个点节点和N个层节点，然后按照层与层之间（双向，权值C）、点与点之间（即后来给的M条边）、点与相对应的层之间（层指向点，权值0），点与对应层的相邻层之间（点指向层，权值C）建图，最后求最短路即可

解释一个代码中难理解的地方：

这里是点和所在层建立关系不能建双向边的原因是假设有两个点在同一层

比如有三个点，点1在第一层，点2也在第一层，虚拟第一层为点4，那么1-4有一条距离为0的点，4-1有一条距离为0的点

2-4有一条距离为0的点,4-2有一条距离为0的点,那么1-2距离就成为0了，这是不对的。

这两个if建立单向边的原因是，如果三层，中间一层没有点，建立双向边会导致最上和最下的两层可以相通，而事实上是不通的

【代码】

#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=2e5+5;
#define pir pair<int,int>
int n,m,c,cas,cnt,S,la[N],dis[N];bool has[N];
struct node{int v,w,next;}e[N<<2];int tot,head[N];bool vis[N];
inline void addedge(int x,int y,int z){
	e[++tot].v=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
}
inline void add(int x,int y,int z){
	addedge(x,y,z);
	addedge(y,x,z);
}
inline void Clear(){
	tot=0;
	memset(head,0,sizeof head);
	memset(has,0,sizeof has);
	memset(vis,0,sizeof vis);
	memset(dis,0x3f,sizeof dis);
}
inline void Init(){
	scanf("%d%d%d",&n,&m,&c);
	for(int i=1;i<=n;i++) scanf("%d",&la[i]),has[la[i]]=1;
	for(int i=1,x,y,z;i<=m;i++) scanf("%d%d%d",&x,&y,&z),add(x,y,z);
	for(int i=1;i<=n;i++) if(has[i]&&has[i+1]) add(i+n,i+1+n,c);
	for(int i=1;i<=n;i++){
		addedge(la[i]+n,i,0);
		if(la[i]>1) addedge(i,la[i]+n-1,c);
		if(la[i]<n) addedge(i,la[i]+n+1,c);
	}
}
#define mp make_pair
inline void dijkstra(){
	priority_queue<pir,vector<pir>,greater<pir> >q;
	q.push(mp(dis[S=1]=0,S));//vis[S]=1;
	while(!q.empty()){
		pir t=q.top();q.pop();
		int x=t.second;
		if(vis[x]) continue;
		vis[x]=1;
		for(int i=head[x];i;i=e[i].next){
			int v=e[i].v;
			if(!vis[v]&&dis[v]>dis[x]+e[i].w){
				q.push(mp(dis[v]=dis[x]+e[i].w,v));
			}
		}
	}
	printf("Case #%d: %d\n",++cnt,dis[n]<0x3f3f3f3f?dis[n]:-1);
}
int main(){
	for(scanf("%d",&cas);cas--;){
		Clear();
		Init();
		dijkstra(); 
	}
	return 0;
}

HDU 4725 The Shortest Path in Nya Graph（构图）

The Shortest Path in Nya Graph

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