The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13070 Accepted Submission(s): 2794

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4725

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 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 <= 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

题意：

给出一个分层图，每个点只属于一层，点与点之间到达有一定花费，然后相邻两层移动也有一定花费。最后问从1号点到n号点的最小花费是什么。

题解：

朴素的想法就是点与点之间连边，层与层之间连边，但是空间会爆掉。

于是就像将“层”给分离出去，如果点i属于x层，就往n+x连边，这样会节约很多的空间。

但是这也有一个问题，就是跑最短路的时候点跑向相应的层，然后又跑回来。

所以我们考虑只连出边/入边，然后层直接与点相连。

具体代码如下：

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <iostream>

#include <queue>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

const int N = 2e5+;

int n,m,c,t,tot;

int head[N],d[N],vis[N],belong[N],lay[N];

struct Edge{

    int v,w,next ;

}e[N<<];

struct node{

    int d,u;

    bool operator < (const node &A)const{

        return d>A.d;

    }

};

void adde(int u,int v,int w){

    e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;

}

void Dijkstra(int s){

    priority_queue <node> q;memset(d,INF,sizeof(d));

    memset(vis,,sizeof(vis));d[s]=;

    q.push(node{,s});

    while(!q.empty()){

        node cur = q.top();q.pop();

        int u=cur.u;

        if(vis[u]) continue ;

        vis[u]=;

        for(int i=head[u];i!=-;i=e[i].next){

            int v=e[i].v;

            if(d[v]>d[u]+e[i].w){

                d[v]=d[u]+e[i].w;

                q.push(node{d[v],v});

            }

        }

    }

}

int main(){

    cin>>t;

    int cnt =;

    while(t--){

        cnt++;

        scanf("%d%d%d",&n,&m,&c);

        memset(head,-,sizeof(head));tot=;

        memset(belong,,sizeof(belong));

        memset(lay,,sizeof(lay));

        for(int i=,tmp;i<=n;i++){

            scanf("%d",&tmp);

            lay[tmp]=;

            belong[i]=tmp;

        }

        for(int i=;i<=n;i++){

            if(!lay[i] || !lay[i-]) continue ;

            adde(i+n,i+n-,c);

            adde(i+n-,i+n,c);

        }

        for(int i=;i<=n;i++){

            int tmp = belong[i];

            adde(n+tmp,i,);

            if(tmp>) adde(i,tmp+n-,c);

            if(tmp<n) adde(i,tmp+n+,c);

            //adde(i,n+tmp,0);

        }

        for(int i=,u,v,w;i<=m;i++){

            scanf("%d%d%d",&u,&v,&w);

            adde(u,v,w);adde(v,u,w);

        }

        Dijkstra();

        printf("Case #%d: ",cnt);

        if(d[n]==INF) puts("-1");

        else cout<<d[n]<<endl;

    }

    return ;

}

巴特西

HDU4725：The Shortest Path in Nya Graph（最短路）

The Shortest Path in Nya Graph

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