COGS 315. [POJ3255] Floor tile roadblocks

315. [POJ3255] Floor tile roadblocks

★ ¡ï Input File: block.in output file: block.out Simple comparison
Time limit: 1 s memory limit: MB

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).

Bessie moved to a small farm to live, but she enjoyed the occasional visit to a good friend of hers. Bessie liked the scenery along the way, so she wanted to go as slowly as possible. So she decided to take a short circuit instead of the shortest way.

Suburbs contain R (1<=r<=100,000) Two-way road, they are connected to the N (1<=n<=5000) intersection (node), in order to facilitate us to mark them as 1. N. Bessie departs from node 1th, her good friend (destination) at node N.

The short-circuit and the shortest-circuit can be partially coincident, the secondary short-circuit may also go through the same path or the same node multiple times. The short-circuit is the shortest path in a short circuit. When there are multiple shortest circuits present, the secondary short circuit is the least of all other paths.

Input

Line 1:two space-separated integers: N and R

Lines 2.. R+1:each line contains three space-separated integers: a, B, and D that describe A ro Ad that connects intersections A and B and have length D (1≤ d ≤5000)

Input:

Line 1th: Two integers separated by a space N and R

2nd.. R+1 lines: Each line consists of three integers, a, B, and D, separated by spaces. The length of the path used to describe the connection between Node A and Node B is D (1<=d<=5000).

Output

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

Output:

First line: An integer representing the length of the secondary short circuit from node 1 to 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

Tips:

The route from node 1 to node n in the sample is 1->2->4 (len=100+200=300 (Shortest path)) and 1->2->3->4 (100+250+100=450 (short circuit)) respectively.

Translation BYKZFFFFFFFF

Idea: secondary short circuit board

#include <iostream>#include<queue>#include<cstdio>#include<cstring>#include<algorithm>#defineINF 0x3f3f3f3f#defineMAXN 400000using namespacestd;structnond{intg,f,to; BOOL operator< (ConstNond &r)Const {        if(r.f==f)returnr.g<G; Else returnr.f<F; }}tmp,opt;intN,m,p,s,t,cnt,tot,tot1,tot2;intDIS[MAXN],VIS[MAXN],DIS2[MAXN],VIS2[MAXN];intTO[MAXN],NET[MAXN],CAP[MAXN],HEAD[MAXN];intTO1[MAXN],NET1[MAXN],CAP1[MAXN],HEAD1[MAXN];voidAddintUintVintW) {to[++tot]=v;net[tot]=head[u];cap[tot]=w;head[u]=tot; to[++tot]=u;net[tot]=head[v];cap[tot]=w;head[v]=tot; to1[++tot1]=v;net1[tot1]=head1[u];cap1[tot1]=w;head1[u]=tot1; to1[++tot1]=u;net1[tot1]=head1[v];cap1[tot1]=w;head1[v]=tot1;}voidSPFA (ints) {Queue<int>que1;  for(intI=0; i<=n;i++) dis[i]=INF;    Que1.push (s); Vis[s]=1;d is[s]=0;  while(!Que1.empty ()) {        intnow=Que1.front ();        Que1.pop (); Vis[now]=0;  for(intI=head1[now];i;i=Net1[i])if(dis[to1[i]]>dis[now]+Cap1[i]) {Dis[to1[i]]=dis[now]+Cap1[i]; if(!Vis[to1[i]]) {Vis[to1[i]]=1;                Que1.push (To1[i]); }            }    }}intAstar (intKK) {Priority_queue<nond>que; if(s==t) kk++; if(Dis[s]==inf)return-1; TMP.G=0; Tmp.to=s; TMP.F=Dis[s];    Que.push (TMP);  while(!Que.empty ()) {tmp=Que.top ();        Que.pop (); if(tmp.to==t) cnt++; if(CNT==KK)returnTMP.G;  for(intI=head[tmp.to];i;i=Net[i]) {opt.to=To[i]; OPT.G=tmp.g+Cap[i]; OPT.F=opt.g+Dis[to[i]];        Que.push (opt); }    }    return-1;}intMain () {Freopen ("block.in","R", stdin); Freopen ("Block.out","W", stdout); scanf ("%d%d",&n,&m);  for(intI=1; i<=m;i++){        intA,b,c; scanf ("%d%d%d",&a,&b,&c);    Add (A,B,C); } s=1; t=N;    SPFA (t);  for(intI=1; i<=n;i++) {CNT=0; intans=Astar (i); if(ans!=Dis[s]) {printf ("%d", ans); return 0; }    }}

COGS 315. [POJ3255] Floor tile roadblocks