杭电4725 The Shortest Path in Nya Graph（Dijkstra+堆优化）

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2752    Accepted Submission(s): 658

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

/*重在构图，，，注意的是点与层入度相连，到了下边层与层中间相连的时候要用与点相连的那条边与下一层相连，否则会出错。。。。Time:2015-1-8 8:44*/#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;const int MAX=1000000+10;struct qNode{    int v,c;    qNode(int _v=0,int _c=0):v(_v),c(_c){};    bool operator<(const qNode &a)const{        return c>a.c;    }};struct Edge{    int v,c;    Edge(int _v=0,int _c=0):v(_v),c(_c){};};vector<Edge>E[MAX];void addEdge(int u,int v,int w){    E[u].push_back(Edge(v,w));}int dis[MAX];bool vis[MAX];void Dijkstra(int s,int n){    priority_queue<qNode>pq;    memset(vis,0,sizeof(vis));    while(!pq.empty())pq.pop();    for(int i=1;i<=n;i++)dis[i]=INF;    dis[s]=0;//vis[s]=true;此处不能标记。因为此处表示的v是入队的点s(v)    pq.push(qNode(s,0));    while(!pq.empty()){        qNode tmp=pq.top();pq.pop();        int u=tmp.v;        if(vis[u])continue;            vis[u]=true;        for(int i=0;i<E[u].size();i++){            int v=E[u][i].v;            int cost=E[u][i].c;            if(!vis[v]&&dis[v]>dis[u]+cost){                dis[v]=dis[u]+cost;                pq.push(qNode(v,dis[v]));            }        }    }}int main(){    int T;    int nCase=1;    int n,m,c;    scanf("%d",&T);    while(T--){        scanf("%d%d%d",&n,&m,&c);        for(int i=1;i<=3*n;i++)E[i].clear();//3n        int layer,u,v,w;        for(int i=1;i<=n;i++){            scanf("%d",&layer);            addEdge(i,n+layer*2-1,0);            addEdge(n+layer*2,i,0);        }        for(int i=1;i<n;i++){//i层的入度表示与上边连接点的表示要对应            addEdge(n+i*2-1,n+(i+1)*2,c);            addEdge(n+(i+1)*2-1,n+i*2,c);        }        for(int i=1;i<=m;i++){            scanf("%d%d%d",&u,&v,&w);            addEdge(u,v,w);            addEdge(v,u,w);        }        Dijkstra(1,3*n);        printf("Case #%d: ",nCase++);        if(dis[n]==INF)puts("-1");        else{            printf("%d\n",dis[n]);        }    }return 0;}