ACM学习历程—HDU4725 The Shortest Path in Nya Graph（SPFA && 优先队列）

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 <= 10 ⁵) and C(1 <= C <= 10 ³), 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 <= 10 ⁴), 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

 

题目大意是是有n层。然后每层会有部分点，点与点可以通过特点的代价互达，相邻层之间的点可以通过C代价到达。

题目一开始没有看到每层可以有多个点。还有需要注意的是同层之间的点不能直接0代价到达。

之前在没考虑同层间不能直接0代价互达的时候，是把每个层看作一个节点，这样某层的节点到这个节点的代价都是0。

如果考虑到这个条件的话，

所以需要构造出下面这样的路径:

为每层设置From和To节点。某层的节点只能从From节点前往，或者前往To节点。

而相邻层之间亦是如此。

这样就能避免同层代价为0这个问题了。

为了避免点的冲突，From从n开始往后，To从2n开始往后。

需要注意的是这样操作以后点的数目上界会变成3N,线段就是6N。

这里采用链式前向星存图，采用STL优先队列的SPFA。

 

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <utility>
#include <queue>
#include <vector>
#define N 300005

using namespace std;

typedef pair<int, int> pii;

struct Edge
{
    int to;
    int next;
    int val;
}edge[2*N];

int head[N], cnt;

void addEdge(int u, int v, int w)
{
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    edge[cnt].val = w;
    head[u] = cnt;
    cnt++;
}

int dis[N];
bool vis[N];
int n, m, c;

inline int idFrom(int i)
{
    return i + n;
}

inline int idTo(int i)
{
    return i + 2*n;
}

void Input()
{
    memset(head, -1, sizeof(head));
    memset(dis, -1, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[1] = 0;
    cnt = 0;
    int k, v, w;
    scanf("%d%d%d", &n, &m, &c);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &k);
        addEdge(idFrom(k), i, 0);
        addEdge(i, idTo(k), 0);
    }
    for (int i = 1; i < n; ++i)
    {
        addEdge(idTo(i), idFrom(i+1), c);
        addEdge(idTo(i+1), idFrom(i), c);
    }
    for (int i = 0; i < m; ++i)
    {
        scanf("%d%d%d", &k, &v, &w);
        addEdge(k, v, w);
        addEdge(v, k, w);
    }
}

void Work()
{
    pii k;
    priority_queue <pii, vector<pii>, greater<pii> > q;
    q.push(pii(0, 1));
    while (!q.empty())
    {
        k = q.top();
        q.pop();
        int x = k.second;
        if (vis[x])
            continue;
        vis[x] = true;
        for (int i = head[x]; i != -1; i = edge[i].next)
        {
            if (dis[edge[i].to] != -1 &&
                dis[edge[i].to] <= dis[x] + edge[i].val)
                continue;
            if (vis[edge[i].to])
                continue;
            dis[edge[i].to] = dis[x] + edge[i].val;
            q.push(pii(dis[edge[i].to], edge[i].to));
        }
    }
    printf("%d\n", dis[n]);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        printf("Case #%d: ", times);
        Input();
        Work();
    }
    return 0;
}