[HDU 4725]The Shortest Path in Nya Graph[建图]

题目链接：[HDU 4725]The Shortest Path in Nya Graph[建图]
题意分析：
1到n点间，有边通过代价为w。点又在相应的层上，x层上的点可以到达x + 1和 x -1层上的任意一个点，代价为c。问：点1到点n的最小代价是？无法到达输出『-1』
解题思路：
这题可以在点和层之间连一条边，这里我们设n + 2 * x为层入口，n + 2*x - 1为层出口。这样把层拆开是避免同层上的点之间的距离变为0。
个人感受：
果然图见多了才有感觉，学习了~
具体代码如下：

#include<cstdio>#include<cstring>#include<map>#include<queue>#define pii pair<int, int>using namespace std;const int INF = 0x7f7f7f7f;const int MAXN = 1e5 + 111;struct Node{    int to, w;    Node(int _to, int _w): to(_to), w(_w){}};vector<Node> G[3 * MAXN];int dis[3 * MAXN];void dijkstra(int s){    dis[s] = 0;    priority_queue<pii, vector<pii>, greater<pii> > pq;    pq.push(pii(dis[s], s));    while (pq.size())    {        int cur = pq.top().second, curDis = pq.top().first; pq.pop();        if (curDis > dis[cur]) continue;        for (int i = 0; i < G[cur].size(); ++i)        {            Node &e = G[cur][i];            if (dis[e.to] > dis[cur] + e.w)            {                dis[e.to] = dis[cur] + e.w;                pq.push(pii(dis[e.to], e.to));            }        }    }}int main(){    int t, n, m, c, x;    for (int kase = scanf("%d", &t); kase <= t; ++kase)    {        scanf("%d%d%d", &n, &m, &c);        for (int i = 0; i <= 3 * n; ++i) G[i].clear(), dis[i] = INF;        for (int i = 1; i <= n; ++i)        {            scanf("%d", &x);            G[i].push_back(Node(n + 2 * x, 0));     // 点和入口建边            G[n + 2 * x - 1].push_back(Node(i, 0)); // 出口和点建边        }        for (int i = 1; i <= n; ++i) // 层与层间        {            if (i > 1)                G[n + 2 * i].push_back(Node(n + 2 * (i - 1) - 1, c));            if (i < n)                G[n + 2 * i].push_back(Node(n + 2 * (i + 1) - 1, c));        }        int u, v, w;        for (int i = 0; i < m; ++i)        {            scanf("%d%d%d", &u, &v, &w);            G[u].push_back(Node(v, w));            G[v].push_back(Node(u, w));        }        dijkstra(1);        printf("Case #%d: ", kase);        if (dis[n] == INF) puts("-1");        else printf("%d\n", dis[n]);    }    return 0;}