Tricks Device

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1584 Accepted Submission(s): 388

Problem Description

Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the chambers. Innocent Wu wants to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways from the entrance to the end of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.

Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.

Input

There are multiple test cases. Please process till EOF.

For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.

In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.

The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.

Output

Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.

Sample Input

       8 9 1 2 2 2 3 2 2 4 1 3 5 3 4 5 4 5 8 1 1 6 2 6 7 5 7 8 1 

Sample Output

2 6

Author

FZUACM

Source

/* ***********************************************Author        :CKbossCreated Time  :2015年07月24日 星期五 10时07分09秒File Name     :HDOJ5294.cpp************************************************ */#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               using namespace std;typedef pair
                
                  pII;const int INF=0x3f3f3f3f;const int maxn=2200;int n,m;/*************EDGE********************/struct Edge{ int to,next,cost,cap,flow;}edge[maxn*60],edge2[maxn*60];int Adj[maxn],Size;int Adj2[maxn],Size2;void Add_Edge(int u,int v,int c){ edge[Size].to=v; edge[Size].next=Adj[u]; edge[Size].cost=c; Adj[u]=Size++;}/********spfa************/int dist[maxn];bool inQ[maxn];vector
                 
                   Pre[maxn];int spfa(Edge* edge,int* Adj){ memset(dist,63,sizeof(dist)); memset(inQ,false,sizeof(inQ)); dist[1]=0; queue
                  
                    q; inQ[1]=true;q.push(1); while(!q.empty()) { int u=q.front();q.pop(); for(int i=Adj[u];~i;i=edge[i].next) { int v=edge[i].to; if(dist[v]>dist[u]+edge[i].cost) { Pre[v].clear(); Pre[v].push_back(u); dist[v]=dist[u]+edge[i].cost; if(!inQ[v]) { inQ[v]=true; q.push(v); } } else if(dist[v]==dist[u]+edge[i].cost) { Pre[v].push_back(u); } } inQ[u]=false; } return dist[n];}/********************rebuild************************/void Add_Edge2(int u,int v,int w,int rw=0){ edge2[Size2].cost=1; edge2[Size2].to=v; edge2[Size2].cap=w; edge2[Size2].next=Adj2[u]; edge2[Size2].flow=0; Adj2[u]=Size2++; edge2[Size2].cost=1; edge2[Size2].to=u; edge2[Size2].cap=w; edge2[Size2].next=Adj2[v]; edge2[Size2].flow=0; Adj2[v]=Size2++;}bool used[maxn];int edges;void rebuild(){ memset(used,false,sizeof(used)); queue
                   
                     q; q.push(n); used[n]=true; edges=0; while(!q.empty()) { int v=q.front(); q.pop(); for(int i=0,sz=Pre[v].size();i
                    
                     v //cout<
                     <<" ---> "<
                      
                       <
                       
                        edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow; } for(int i=pre[u];~i;i=pre[edge[i^1].to]) { edge[i].flow+=Min; edge[i^1].flow-=Min; } u=start; ans+=Min; continue; } bool flag=false; int v; for(int i=cur[u];~i;i=edge[i].next) { v=edge[i].to; if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u]) { flag=true; cur[u]=pre[v]=i; break; } } if(flag) { u=v; continue; } int Min=N; for(int i=Adj2[u];~i;i=edge[i].next) { if(edge[i].cap-edge[i].flow&&dep[edge[i].to]

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