POJ question 3255 Roadblocks (Short Circuit)

POJ question 3255 Roadblocks (Short Circuit)
Roadblocks

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:8670		Accepted:3138

Description

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. she does not want to get to her old home too quickly, because she likes the scenery along the way. she has decided to take the second-shortest rather than the shortest path. she knows there must be some second-shortest path.

The countryside consistsR(1 ≤R≤ 100,000) bidirectional roads, each linking two of the N (1 ≤N≤ 5000) intersections, conveniently numbered 1 ..N. Bessie starts at intersection 1, and her friend (the destination) is at intersectionN.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack I. e ., use the same road or intersection more than once. the second-shortest path is the shortest path whose length is longer than the shortest path (s) (I. e ., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path ).

Input

Line 1: Two space-separated integers: NAnd R
Lines 2 .. R+ 1: Each line contains three space-separated integers: A, B, And DThat describe a road that connects intersections AAnd BAnd has length D(1 ≤ D≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 41 2 1002 4 2002 3 2503 4 100

Sample Output

450

Hint

Two routes: 1-> 2-> 4 (length 100 + 200 = 300) and 1-> 2-> 3-> 4 (length 100 + 250 + 100 = 450)

Source

USACO 2006 November Gold

Ac code

 

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std;#define INF 0xfffffffint n,m;struct s{int u,v,w,next;}edge[200100];int head[5050],cnt,disr[5050],dis[5050],vis[5050],ans;void add(int u,int v,int w){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;}void init(){int i;for(i=0;i<=n;i++){dis[i]=INF;disr[i]=INF;}}void spfa(int s,int *dis){queue
     
      q;memset(vis,0,sizeof(vis));dis[s]=0;vis[s]=1;q.push(s);while(!q.empty()){int u=q.front();q.pop();vis[u]=0;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;int w=edge[i].w;if(dis[v]>dis[u]+w){dis[v]=dis[u]+w;if(!vis[v]){vis[v]=1;q.push(v);}}}}}int main(){while(scanf("%d%d",&n,&m)!=EOF){int i,j;memset(head,-1,sizeof(head));cnt=0;init();for(i=0;i
      
       dis[n]&&temp