HDU 4725 The Shortest Path in Nya Graph （最短路拆点建图）

最新推荐文章于 2021-09-30 20:35:16 发布

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最新推荐文章于 2021-09-30 20:35:16 发布

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分类专栏： *Graph Theory

本文链接：https://blog.csdn.net/Codeblocksm/article/details/51458503

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本文介绍了一种求解特殊图结构中最短路径的问题，通过构建双倍节点数目的图，并利用Dijkstra算法优化来解决从起点到终点的最短路径问题。

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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 <= 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 li (1 <= li <= N), which is the layer of ith 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

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。

到第i层的入边是点N+2*i-1, 出边从点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i+1层。

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

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

保证边建对。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

const int N = 1e5 + 100;
const int MAXN = 3*N;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int v,w;
    Edge(int v,int w):v(v),w(w){}
};
vector<Edge> E[MAXN]; int dis[MAXN]; bool vis[MAXN];
void add(int u,int v,int w)
{
    E[u].push_back(Edge(v,w));
};
struct Node
{
    int v,w;
    Node(){}
    Node(int v,int w):v(v),w(w){}
    friend bool operator < (const Node &a,const Node &b)
    {
        return a.w > b.w;
    }
};
int dijstra(int n)
{
    for(int i=1;i<=3*n;i++) dis[i] = INF;
    memset(vis,false,sizeof(vis));
    priority_queue<Node>que;
    que.push(Node(1,0));
    dis[1] = 0;
    while(!que.empty())
    {
        Node now = que.top();
        que.pop();

        int u = now.v;
        if(vis[u]) continue;
        vis[u] = true;

        for(int i=0;i<E[u].size();i++)
        {
            int v = E[u][i].v;
            int c = E[u][i].w;
            if(!vis[v] && dis[v] > dis[u] + c)
            {
                dis[v] = dis[u] + c;
                que.push(Node(v,dis[v]));
            }
        }
    }
    if(dis[n] == INF) dis[n] = -1;
    return dis[n];
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas = 1; cas<=T ;cas++)
    {
        int n,m,c;
        scanf("%d%d%d",&n,&m,&c);
        for(int i=1;i<=3*n;i++) E[i].clear();
        for(int i=1;i<=n;i++)
        {
            int x; scanf("%d",&x);
            add(n+2*x-1,i,0);
            add(i,n+2*x,0);
        }
        for(int i=1;i<n;i++)
        {
            add( n+2*(i+1),n+2*i-1,c);
            add( n+2*i,n+2*(i+1)-1,c);
        }
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        printf("Case #%d: %d\n",cas,dijstra(n));
    }
    return 0;
}