hdu 4725 The Shortest Path in Nya Graph

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 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 <= 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

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 100005
#define inf 0x3f3f3f3f
using namespace std;
int n,m,C;
struct Edges
{
    int x,y,w,next;
};
struct HeapNode
{
    int d;
    int u;
    bool operator < (const HeapNode& rhs) const{
        return d>rhs.d;
    }
};
priority_queue<HeapNode>Q;
Edges e[N*10];
int head[3*N];
int dis[3*N];
bool done[3*N];
int k;

void AddEdge(int x,int y,int w)
{
    e[k].x=x,e[k].y=y,e[k].w=w,e[k].next=head[x],head[x]=k++;
}
int DJ()
{
    memset(dis,inf,sizeof(dis));
    memset(done,false,sizeof(done));
    dis[1]=0;
    HeapNode p;
    p.d=dis[1];
    p.u=1;
    Q.push(p);
    while(!Q.empty())
    {
        HeapNode x = Q.top();
        Q.pop();
        int u=x.u;
        if(done[u])
        continue;
        done[u]=true;
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int y=e[i].y;
            int w=e[i].w;
            if(dis[u]+w<dis[y])
            {
                dis[y]=dis[u]+w;
                HeapNode temp;
                temp.d=dis[y];
                temp.u=y;
                Q.push(temp);
            }
        }
    }
    return dis[n];
}

int main()
{
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&C);
        memset(head,-1,sizeof(head));
        k=0;
        for(int i=1;i<=n;i++)
        {
            int num;
            scanf("%d",&num);
            AddEdge(i,n+2*num-1,0);
            AddEdge(n+2*num,i,0);
        }
        for(int i=1;i<n;i++)
        {
            AddEdge(n+2*i-1,n+2*(i+1),C);
            AddEdge(n+2*(i+1)-1,n+2*i,C);
        }
        for(int i=0;i<m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            AddEdge(x,y,z);
            AddEdge(y,x,z);
        }
        int res=DJ();
        if(res==inf)
        {
            printf("Case #%d: -1\n",cas++);
        }
        else
        {
            printf("Case #%d: %d\n",cas++,res);
        }
    }
    return 0;
}

最短路问题，主要考的是构图的思想；

思路：

将每一层拆分成两个点，例如，第一层分成n+1即(n+2*i-1)和n+2(n+2*i)；总共3*n个点；

也就是第i层分成n+2*i-1和n+2*i;规定n+2*i-1始终作为入边，n+2*i始终作为出边;

如果某个点属于第i层，就连边i->n+2*i-1,n+2*i->i，权值为0;

然后根据题意，相邻的层次连边，权值为C，n+2*i-1->n+2*(i+1)和n+2*(i+1)-1->n+2*i;

然后跟着输入的边构图；

基本上这题的图构好之后，跑一边迪杰斯特拉+优先队列的模板，就能解决此题；

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