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

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

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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。。

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#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;}