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

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 37    Accepted Submission(s): 6

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

 

最短路。

主要是建图。

N个点，然后有N层，要假如2*N个点。

总共是3*N个点。

点1~N就是对应的实际的点1~N.  要求的就是1到N的最短路。

然后点N+1 ~ 3*N 是N层拆出出来的点。

第i层，入边到N+2*i-1, 出边从N+2*i 出来。（1<= i <= N）

N + 2*i    到  N + 2*(i+1)-1 加边长度为C. 表示从第i层到第j层。

N + 2*(i+1) 到 N + 2*i - 1 加边长度为C,表示第i+1层到第j层。

如果点i属于第u层，那么加边 i -> N + 2*u -1         N + 2*u ->i  长度都为0

然后用优先队列优化的Dijkstra就可以搞出最短路了

/* ***********************************************
Author        :kuangbin
Created Time  :2013-9-11 12:30:12
File Name     :2013-9-111010.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

/*
 * 使用优先队列优化Dijkstra算法
 * 复杂度O(ElogE)
 * 注意对vector<Edge>E[MAXN]进行初始化后加边
 */
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 Dijkstra(int n,int start)//点的编号从1开始
{
    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 addedge(int u,int v,int w)
{
    E[u].push_back(Edge(v,w));
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    int N,M,C;
    scanf("%d",&T);
    int iCase = 0;
    while(T--)
    {
        scanf("%d%d%d",&N,&M,&C);
        for(int i = 1;i <= 3*N;i++) E[i].clear();
        int u,v,w;
        for(int i = 1;i <= N;i++)
        {
            scanf("%d",&u);
            addedge(i,N + 2*u - 1,0);
            addedge(N + 2*u ,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);
        }
        while(M--)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        Dijkstra(3*N,1);
        iCase++;
        if(dist[N] == INF)dist[N] = -1;
        printf("Case #%d: %d
",iCase,dist[N]);

    }
    return 0;
}