Secondary short Circuit

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7921		Accepted: 2896

Description

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

The countryside consists of R (1≤ R ≤100,000) bidirectional roads, each linking both of the N (1≤ N ≤5000) intersections, conveniently numbered 1.. n. Bessie starts at intersection 1, and she friend (the destination) is at intersection N.

The Second-shortest path may share roads with any of the shortest paths, and it could backtrack i.e., use the same road or I Ntersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path (s) (i.e., if EST paths exist, the second-shortest path is the one whose length are 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's 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

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

At first I thought just for the end of the N to find a second short-circuit, only apply for a DIS2 variable, and then update Dis[n] When the DIS2 can be updated, but the "program design" This book is good, not only to find the end of the second short circuit, but also to find all the nodes of the short-circuit, so there is a dis2[ MAXN].

#define_crt_secure_no_deprecate#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<functional>using namespacestd;#defineMAXR 100001#defineMAXN 5002#defineINF 100000000intN, R;//intersections,roadsstructedge{intto, cost; Edge (Const intAConst intb): to (a), cost (b) {}};vector<edge> G[MAXN];//adjacency Matrixtypedef pair<int,int>P;priority_queue<p, Vector<p&gt, greater<p>> que;//Priority Queue DijkstraintDIS[MAXN], DIS2[MAXN];voidsolve () {//initializationFill (dis, dis +MAXN, INF); Fill (Dis2, Dis2+MAXN, INF); //CIN >> N >>R; intA, B, C;  while(r--) {scanf (" %d%d%d", &a, &b, &c);　　G[a].push_back (Edge (b, c));    At first I just applied for the edge temp variable, which will be wrong when called, so I added the constructor G[b].push_back (Edge (A, c)) directly; }    //shortest way to start from N=1dis[1] =0; Que.push (P (0,1));  while(!Que.empty ()) {P PIR=que.top (); Que.pop (); intv = pir.second, d =Pir.first; if(Dis2[v] < D)Continue;  for(inti =0; I < g[v].size (); i++) {Edge&e =G[v][i]; intD2 = E.cost +D; if(dis[e.to]>D2)                {Swap (dis[e.to], D2);            Que.push (P (dis[e.to], e.to)); }            if(Dis2[e.to]>d2&&dis[e.to] <D2) {Dis2[e.to]=D2;            Que.push (P (dis2[e.to], e.to)); }}} printf ("%d\n", Dis2[n]);}intMain () {solve ();//GetChar ();    return 0;}

Secondary short Circuit