HDU 4725 最短路构图

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1319    Accepted Submission(s): 293

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

构造虚拟节点，然后见图，dijstra优先队列求最短路。

代码：

 

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<map>
#include<list>
#include<set>
#include<queue>
#include<stack>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=1000010;
struct qnode
{
    int v;
    int c;
    qnode(int _v=0,int _c=0):v(_v),c(_c){}
    bool operator <(const qnode &r)const
    {
        return c>r.c;
    }
};
struct Edge
{
    int v,cost;
    Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
bool vis[MAXN];
int dist[MAXN];
void fun(int n,int start)
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)dist[i]=INF;
    priority_queue<qnode>que;
    while(!que.empty())que.pop();
    dist[start]=0;
    que.push(qnode(start,0));
    qnode tmp;
    while(!que.empty())
    {
        tmp=que.top();
        que.pop();
        int u=tmp.v;
        if(vis[u])continue;
        vis[u]=true;
        for(int i=0;i<E[u].size();i++)
        {
            int v=E[tmp.v][i].v;
            int cost=E[u][i].cost;
            if(!vis[v]&&dist[v]>dist[u]+cost)
            {
                dist[v]=dist[u]+cost;
                que.push(qnode(v,dist[v]));
            }
        }
    }
}
void add(int u,int v,int w)
{
    E[u].push_back(Edge(v,w));
}
int main()
{
        int i,j,k,m,t,T,C,p,q,g,h,u,v,w,n;
        scanf("%d",&T);
        for(t=1;t<=T;t++)
        {
                scanf("%d%d%d",&n,&m,&C);
             for( i = 1;i <= 3*n;i++) E[i].clear();
                for( i = 1;i <= n;i++)
        {
            scanf("%d",&u);
            add(i,n + 2*u - 1,0);
            add(n + 2*u ,i,0);

        }
        for( i = 1;i < n;i++)
        {
            add(n + 2*i-1,n + 2*(i+1),C);
            add(n + 2*(i+1)-1,n + 2*i,C);
        }
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
                printf("Case #%d: ",t);
                for(i=1;i<=3*n;i++)dist[i]=INF;
                fun(3*n,1);
                int ans;
                if(dist[n]>=INF)ans=-1;
                else ans=dist[n];
                printf("%d\n",ans);
        }
        return 0;
}