hdu 4725 The Shortest Path in Nya Graph(最短路)

The Shortest Path in Nya Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3657    Accepted Submission(s): 852

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

 

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zhuyuanchen520

 

题意：有n个点 , m条无向边 ,每条边都是有权值, 并且每个点属于一个楼层 ,楼层上的点到可以到相邻楼层的任意一点 ,但是要花费 c 。

      没有点的相邻楼层不能互达。求 1 到 n的最小花费。

题解：图建好了就是裸的最短路了。但是建图有点麻烦，参考了别人的代码之后才明白为什么要这样建图。

      把楼层看成一个点，第i层可以看成第n+i个点。楼层与该楼层上的点建边，边权为0，单向；楼层与

      相邻楼层建边，边权为C，双向；相邻楼层上的点与该楼层建边，边权为C，单向。

#include<cstring>#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#define N 200100#define ll long longusing namespace std;const int inf=0x3f3f3f3f;int n,m,c;int lay[N];bool hava[N];bool vis[N];int cnt[N];int dist[N];struct Edge {    int v;    int cost;    Edge(int _v=0,int _c=0):v(_v),cost(_c) {}    bool operator <(const Edge &r)const {        return cost>r.cost;    }};vector<Edge>E[N];void addedge(int u,int v,int w) {    Edge it;    it.v=v;    it.cost=w;    E[u].push_back(it);}void Dijk(int n,int start) { //点的编号从1开始    memset(vis,false,sizeof(vis));    for(int i=1; i<=n*2; i++)dist[i]=inf;    priority_queue<Edge>que;    while(!que.empty())que.pop();    dist[start]=0;    que.push(Edge(start,0));    Edge tmp;    while(!que.empty()) {        tmp=que.top();        que.pop();        int u=tmp.v;        if(vis[u])continue;        vis[u]=true;        for(int i=0; i<E[u].size(); i++) {            int v=E[tmp.v][i].v;            int cost=E[u][i].cost;            if(!vis[v]&&dist[v]>dist[u]+cost) {                dist[v]=dist[u]+cost;                que.push(Edge(v,dist[v]));            }        }    }}int main() {   // freopen("test.in","r",stdin);    int t;    cin>>t;    int ca=1;    while(t--) {        for(int i=0; i<=n*2+1; i++)E[i].clear();        scanf("%d%d%d",&n,&m,&c);        memset(hava,0,sizeof hava);        for(int i=1; i<=n; i++) {            scanf("%d",&lay[i]);            hava[lay[i]]=true;        }        int u,v,cost;        for(int i=1; i<=m; i++) {            scanf("%d%d%d",&u,&v,&cost);            addedge(u,v,cost);            addedge(v,u,cost);        }        if(n<=1) {            printf("Case #%d: 0\n",ca++);            continue;        }        for(int i=1; i<n; i++) {            if(hava[i]&&hava[i+1]) {                addedge(n+i,n+i+1,c);                addedge(n+1+i,n+i,c);            }        }        for(int i=1; i<=n; i++) {            addedge(lay[i]+n,i,0);            if(lay[i]>1)addedge(i,lay[i]-1+n,c);            if(lay[i]<n)addedge(i,lay[i]+1+n,c);        }        Dijk(n,1);        printf("Case #%d: %d\n",ca++,dist[n]>=inf?-1:dist[n]);    }    return 0;}