HDU 4725 The Shortest Path in Nya Graph（優先隊列優化 Dijkstra+建圖）

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 <= 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

23 3 31 3 21 2 12 3 11 3 33 3 31 3 21 2 22 3 21 3 4  

Sample Output

Case #1: 2Case #2: 3

題意：

第一行給出n，m，c，分别表示n個點，m條無向邊，樓層之間的花費（可跳躍至相鄰樓層中任意點）。

接下來一行n個數字，分别表示 每點得歸屬樓層。

接下來給出m行，無向邊u，v，w，w代表權值。

求1到n得最小權值

思路：

點與點之間的路線比較好解決（直接建邊）。

點與層之間建立如下邊：

點 到 該點所在層出口，點所在層 到該點，各個出口到其相鄰入口。（這樣建圖就不會，因為2點所在層一樣犯瞬移的錯誤，和無法連跳兩層的錯誤）

最短路采用 Dijkstra+優先隊列 方法

代碼：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

#define For(a,b,c) for(int a = b; a <= c; a++)
#define mem(a,b) memset(a,b,sizeof(a))
int first[300005], tot;

struct Node
{
    int v, w, next;
}e[600005];

struct Point
{
    int dis, x;
    Point(int a = 0, int b = 0): x(a), dis(b){}
    friend bool operator <(Point a, Point b)
    {
        return a.dis > b.dis;
    }
};

void add(int u, int v, int w)
{
    e[tot].v = v;
    e[tot].w = w;
    e[tot].next = first[u];
    first[u] = tot++;
}

int dis[300005];
bool book[300005];

int Dijkstra(int u, int n)
{
    mem(book,false);
    mem(dis,0x3f);
    dis[u] = 0;
    priority_queue<Point> q;
    q.push(Point(u,0));

    Point st;
    while(!q.empty())
    {
        st = q.top();
        q.pop();
        if(book[st.x]) continue;
        book[st.x] = true;
        for(int i = first[st.x]; ~i; i = e[i].next)
        {
            if(dis[e[i].v] > dis[st.x] + e[i].w)
            {
                dis[e[i].v] = dis[st.x] + e[i].w;
                q.push( Point(e[i].v, dis[e[i].v]) );
            }
        }
    }
    if(dis[n] == 0x3f3f3f3f) return -1;
    return dis[n];
}

int main()
{
    int T, cnt = 1;
    scanf("%d",&T);
    while(T--)
    {
        int N, M , C;
        scanf("%d%d%d",&N,&M,&C);
        mem(first,-1);
        tot = 0;
        int u, v, w;

        For(i,1,N)
        {
            scanf("%d",&u);
            add(i,u+N,0);//i->i的出口
            add(u+2*N,i,0);//i的入口->i
        }

        For(i,1,N)
        {
            //出口到相鄰入口
            if(i < N) add(N+i,2*N+i+1,C);
            if(i > 1) add(N+i,2*N+i-1,C);
        }

        while(M--)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }

        printf("Case #%d: %d\n",cnt++,Dijkstra(1,N));
    }
    return 0;
}