HDU 4725 The Shortest Path in Nya Graph 建图加SPFA

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4143    Accepted Submission(s): 965

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 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

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

 

题意：告诉几个层次，每个层次中又有一些点，n个点，告诉每个点所在的层，可能有的层没有点，从一层到另一层需要花费C能量，然后M行，表示两个点之间互相到达的能量花费，求从1到n最少要花费多少能量。

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define INF 0x3f3f3f3fconst int N = 200009;int dis[N],a[N],head[N],vis[N],tip[N];struct Node{    int v,w,nt;}f[N*10];//数组开小T半天。int cnt;queue<int>q;void add(int u,int v,int w){    f[cnt].v=v;    f[cnt].w=w;    f[cnt].nt=head[u];    head[u]=cnt++;}void spfa(int n){    int top=0;    q.push(1);    vis[1]=1;    dis[1]=0;    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=f[i].nt)        {            int v=f[i].v;            if(dis[v]>dis[u]+f[i].w)            {                dis[v]=dis[u]+f[i].w;                if(vis[v]==0)                {                    vis[v]=1;                    q.push(v);                }            }        }    }}int main(){    int n,m,c;    int u,v,w;    int T;    scanf("%d",&T);    for(int ca=1;ca<=T;ca++)    {        scanf("%d%d%d",&n,&m,&c);        cnt=0;        memset(dis,INF,sizeof dis);        for(int i=1;i<=2*n;i++)        {            vis[i]=0;            head[i]=-1;            tip[i]=0;        }        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            tip[a[i]]=1;        }        for(int i=1;i<n;i++)//层抽象出来编号分别为n+1 ~ n+n            if(tip[i] && tip[i+1])            {                add(i+n,i+1+n,c);//层与层之间建立联系能量耗费为C                add(i+1+n,i+n,c);            }        for(int i=1;i<=n;i++)        {            add(a[i]+n,i,0);//每层和该层的点之间建立联系，能量耗费为0            if(a[i]>1) {add(i,a[i]+n-1,c);}//每层的点和相邻层之间建立联系            if(a[i]<n) {add(i,a[i]+n+1,c);}        }        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);            add(v,u,w);        }        spfa(n);        if(dis[n]<INF)            printf("Case #%d: %d\n",ca,dis[n]);        else        printf("Case #%d: %d\n",ca,-1);    }    return 0;}