HDU 4725 The Shortest Path in Nya Graph（最短路 SPFA 建图）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4725

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

题意：

n个点，m条边，以及相邻层之间移动的代价c，给出每个点所在的层数，以及m条边，每条边有u，v，c，表示从节点u到v（无向），并且移动的代价 c ，问说从 1 到 n 的代价最小是多少。

PS:

将层抽象出来成为n个点（对应编号依次为n+1 ~ n+n），然后层与层建边，点与点建边，层与在该层上的点建边 （边长为0），点与相邻层建边 （边长为c）。

代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define N 200017
int n, m, k, c;
int Edgehead[N], dis[N];
int vv[N], lay[N];
struct
{
    int v,w,next;
} Edge[20*N];
bool vis[N];
int cont[N];

void init()
{
    memset(Edgehead,0,sizeof(Edgehead));
    memset(vv,0,sizeof(vv));
}

void Addedge(int u,int v,int w)
{
    Edge[k].next = Edgehead[u];
    Edge[k].w = w;
    Edge[k].v = v;
    Edgehead[u] = k++;
}
int SPFA( int start)
{
    queue<int>Q;
    while(!Q.empty()) Q.pop();
    for(int i = 1 ; i <= N ; i++ )
        dis[i] = INF;
    dis[start] = 0;
    //++cont[start];
    memset(vis,false,sizeof(vis));
    vis[start] = 1;
    Q.push(start);
    while(!Q.empty())//直到队列为空
    {
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for(int i = Edgehead[u] ; i!=-1 ; i = Edge[i].next)//注意
        {
            int v = Edge[i].v;
            int w = Edge[i].w;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u]+w;
                if( !vis[v] )//防止出现环，也就是进队列重复了
                {
                    Q.push(v);
                    vis[v] = true;
                }
                //if(++cont[v] > n)//有负环
                //   return -1;
            }
        }
    }
    return dis[n];
}
int main()
{
    int u, v, w;
    int t;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d%d%d",&n,&m,&c);
        k = 1;
        memset(Edgehead,-1,sizeof(Edgehead));
        
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&u);//i 在第u层
            lay[i] = u;
            vv[u] = 1;
        }
        
        for(int i = 1; i < n; i++)
        {
            if(vv[i] && vv[i+1])  //两层都出现过点相邻层才建边
            {
                Addedge(n+i,n+i+1,c);
                Addedge(n+i+1,n+i,c);
            }
        }

        for(int i = 1; i <= n; i++)  //层到点建边  点到相邻层建边
        {
            Addedge(n+lay[i],i,0);
            if(lay[i] > 1)
                Addedge(i,n+lay[i]-1,c);
            if(lay[i] < n)
                Addedge(i,n+lay[i]+1,c);
        }

        for(int i = 1 ; i <= m ; i++ )
        {
            scanf("%d%d%d",&u,&v,&w);
            Addedge(u,v,w);//双向链表
            Addedge(v,u,w);//双向链表
        }
        int s = SPFA(1);//从点1开始寻找最短路
        if(s == INF)
        {
            s = -1;
        }
        printf("Case #%d: %d\n",++cas,s);
    }
    return 0;
}