hdu 4725 The Shortest Path in Nya Graph （最短路 两种做法 1.建图+SPFA 2.建图+优先队列+Dijkstra）
The Shortest Path in Nya Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1235 Accepted Submission(s): 274
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.
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.
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.
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
Source
2013 ACM/ICPC Asia Regional Online —— Warmup2
思路：
这题主要难在建图上，要将层抽象出来成为n个点（对应编号依次为n+1~n+n），然后层与层建边，点与点建边，层与在该层上的点建边（边长为0），点与相邻层建边（边长为c）。
ps：这样处理就不用拆点了。不过要注意的是相邻两层必须都要有点才建边（不然会WA，可以参考我贴的数据）。
代码1：SPFA 265MS 9600K
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#define maxn 200005
#define INF 0x3f3f3f3f
using namespace std;
int n,m,c,ans,cnt;
bool vis[maxn],vv[maxn];
int dist[maxn],pp[maxn],lay[maxn];
struct Node
{
int v,w;
int next;
} edge[20*maxn];
queue<int>q;
void init()
{
memset(pp,0,sizeof(pp));
memset(vv,0,sizeof(vv));
memset(dist,0x3f,sizeof(dist));
}
void addedge(int u,int v,int w)
{
cnt++;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=pp[u];
pp[u]=cnt;
}
void SPFA()
{
int i,j,u,v,w;
memset(vis,0,sizeof(vis));
while(!q.empty()) q.pop();
dist[1]=0;
vis[1]=1;
q.push(1);
while(!q.empty())
{
u=q.front();
q.pop();
vis[u]=0;
for(i=pp[u]; i; i=edge[i].next)
{
v=edge[i].v;
w=edge[i].w;
if(dist[v]>dist[u]+w)
{
dist[v]=dist[u]+w;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
}
int main()
{
int i,j,t,u,v,w,test=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&c);
init();
cnt=0;
for(i=1; i<=n; i++)
{
scanf("%d",&u);
lay[i]=u;
vv[u]=1;
}
for(i=1; i<n; i++)
{
if(vv[i]&&vv[i+1]) // 两层都出现过点相邻层才建边
{
addedge(n+i,n+i+1,c);
addedge(n+i+1,n+i,c);
}
}
for(i=1; i<=n; i++) // 层到点建边 点到相邻层建边
{
addedge(n+lay[i],i,0);
if(lay[i]>1) addedge(i,n+lay[i]-1,c);
if(lay[i]<n) addedge(i,n+lay[i]+1,c);
}
for(i=1; i<=m; i++) // 点到点建边
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
SPFA();
printf("Case #%d: ",++test);
ans=dist[n];
if(ans<INF) printf("%d\n",ans);
else printf("-1\n");
}
return 0;
}
/*
1
3 0 1
1 1 1
ans: -1
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 200005
#define INF 0x3f3f3f3f
using namespace std;
int n,m,c,ans,cnt;
bool vis[maxn],vv[maxn];
int dist[maxn],pp[maxn],lay[maxn];
struct node
{
int pos,dd;
bool operator <(const node&xx)const
{
return dd>xx.dd;
}
}cur,now;
struct Node
{
int v,w;
int next;
} edge[20*maxn];
priority_queue<node>q;
void init()
{
memset(pp,0,sizeof(pp));
memset(vv,0,sizeof(vv));
memset(dist,0x3f,sizeof(dist));
}
void addedge(int u,int v,int w)
{
cnt++;
edge[cnt].v=v;
edge[cnt].w=w;
edge[cnt].next=pp[u];
pp[u]=cnt;
}
void Dijkstra()
{
int i,j,u,v,w;
memset(vis,0,sizeof(vis));
while(!q.empty()) q.pop();
dist[1]=0;
cur.pos=1,cur.dd=0;
q.push(cur);
while(!q.empty())
{
now=q.top();
q.pop();
u=now.pos;
if(vis[u]) continue ;
vis[u]=1;
for(i=pp[u]; i; i=edge[i].next)
{
v=edge[i].v;
w=edge[i].w;
if(!vis[v]&&dist[v]>dist[u]+w)
{
dist[v]=dist[u]+w;
cur.pos=v,cur.dd=dist[v];
q.push(cur);
}
}
}
}
int main()
{
int i,j,t,u,v,w,test=0;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&c);
init();
cnt=0;
for(i=1; i<=n; i++)
{
scanf("%d",&u);
lay[i]=u;
vv[u]=1;
}
for(i=1; i<n; i++)
{
if(vv[i]&&vv[i+1])
{
addedge(n+i,n+i+1,c);
addedge(n+i+1,n+i,c);
}
}
for(i=1; i<=n; i++)
{
addedge(n+lay[i],i,0);
if(lay[i]>1) addedge(i,n+lay[i]-1,c);
if(lay[i]<n) addedge(i,n+lay[i]+1,c);
}
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
Dijkstra();
printf("Case #%d: ",++test);
ans=dist[n];
if(ans<INF) printf("%d\n",ans);
else printf("-1\n");
}
return 0;
}