CUGB图论专场：K - The Shortest Path in Nya Graph（dijkstra优先队列优化+线性构图入边）

K - The Shortest Path in Nya Graph

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 
      
      
      
      

初解：刚开始看到这题，好像和B题差不多嘛。没想到敲的代码竟然不对。然后又仔细看子题意，说是相邻层可以用C到达，然后就改了还是不对。看来还是题意不理解啊。把这题目的英语都翻译得自己都全懂了，可是为啥还是不对呢。无语死了……无奈看了别人的题解。原来点与点、层与层之间都得入边……唉……还是自己太嫩了……这种建图以前没有遇到过，搞得现在一头雾水滴！！隐含的建图方式自己还是不会，虽然看别人的题解看懂了，但是这种构图方式看来得多练练了！

思路：

1、对于每个点.向其层的出点做边.距离为0..其层的入点向其做边.距离为0....规定n+2*i为第i层入点,n+2*(i+1)是第i层出点

2、对于每一层..其出点向其相邻层做边.距离为C..

 3、根据题目所给的另外m段...两两间做边...

这样所构的图就可以在线性时间内完成最短路的查找了。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define M 1000002
#define INF 168430090
using namespace std;
typedef long long ll;
int n,m,c,i,u,v,w,tot,ans=INF,head[M],dist[M],vis[M];
struct node
{
    int v,w;
    bool operator < (const struct node a)const
    {
        return a.w < w; //自定义比较函数
    }
}s,x,y;
struct edge
{
    int u,v,w,next;
} e[M];
void add(int u,int v,int w)
{
    e[tot].u = u; e[tot].v = v; e[tot].w = w;
    e[tot].next = head[u]; head[u] = tot++;
}
int dijkstra()
{
    priority_queue<node>q;
    mem(vis,0);
    mem(dist,-1); //也可以初始化为无穷大，最后判断-1情况就得了
    s.v = 1; dist[1]=s.w = 0;
    q.push(s);
    while(!q.empty())
    {
        x = q.top(); q.pop();
        u = x.v;
        if(vis[u]) continue;
        vis[u]=1;
        for(i=head[u]; i!=-1; i=e[i].next)
        {
            v = e[i].v; w = e[i].w;
            if(dist[v]==-1||dist[v]>dist[u]+w)
            {
                dist[v]=dist[u]+w;
                y.v=v; y.w=dist[v];
                q.push(y);
            }
        }
    }
    return dist[n];
}
int main()
{
    int t,k,a;
    sca(t);
    for(k=1;k<=t;k++)
    {
        mem(head,-1); tot = 0;
        scanf("%d%d%d",&n,&m,&c);
        for(i=1;i<=n;i++)
        {
            sca(a);  //结点入边好难
            add(i,n+2*a,0); add(n+2*a+1,i,0);
        }
        for(i=1;i<=n;i++)
        {  //层之间入边也挺难的
            add(n+2*i,n+2*(i+1)+1,c);
            add(n+2*(i+1),n+2*i+1,c);
        }
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w); add(v,u,w);
        }
        printf("Case #%d: %d\n",k,dijkstra());
    }
    return 0;
}