P3003 [Usaco10dec] apple delivery apples Delivery

Title Description

Bessie has a crisp red apples to deliver to both of her friends in the herd. Of course, she travels the C (1 <= c <= 200,000)

Cowpaths which is arranged as the usual graph which connects P (1 <= p <= 100,000) pastures conveniently numbered F Rom 1..p:no Cowpath leads from a pasture to itself, cowpaths is bidirectional, each cowpath have an associated distance, And, best of all, it's always possible-get from any pasture to any other pasture. Each cowpath connects, differing pastures p1_i (1 <= p1_i <= p) and p2_i (1 <= p2_i <= p) with a distance b Etween them of d_i. The sum of the distances d_i does not exceed 2,000,000,000.

What's the minimum total distance Bessie must travel to deliver both apples by starting at pasture PB (1 <= PB <= P ) and visiting Pastures PA1 (1 <= PA1 <= p) and PA2 (1 <= PA2 <= p) in any order. All three of these pastures is distinct, of course.

Consider this map of bracketed pasture numbers and cowpaths with distances:

               3        2       2           [1]-----[2]------[3]-----[4]             \     / \              /             7\   /4  \3           /2               \ /     \          /               [5]-----[6]------[7]                    1       2

If Bessie starts at pasture [5] and delivers apples to pastures [1] and [4], she best path is:

5, 6-> 7, 4, 3 , 2, 1

With a total distance of 12.

Bessie had two sweet and crisp red apples to give to her two friends. Of course, she can go. The C (1<=c<=200000) "Cattle Road" is contained in a common figure, containing the P (1<=p<=100000) pasture, labeled 1, respectively. P. No cattle road will go back to itself from a pasture. "Cattle Road" is two-way, each cattle road will be marked a distance. Most importantly, every ranch can lead to another ranch. Each cattle road is connected to two different pastures p1_i and p2_i (1<=p1_i,p2_i<=p), and the distance is d_i. The sum of the distances of all cattle roads is not greater than 2000000000.

Now, Bessie will start from the ranch PB to Pa_1 and pa_2 Ranch each to send an apple (pa_1 and pa_2 order can be swapped), then the shortest distance is how much? Of course, PB, pa_1 and pa_2 are different.

Input/output format

Input format:

• Line 1:line 1 contains five space-separated integers:c, P, PB, PA1, and PA2

• Lines 2..c+1:line i+1 describes Cowpath I by naming II pastures it connects and the distance between Them:p1_i, P2_i, D _i

Output format:

• Line 1:the Shortest distance Bessie must travel to deliver both apples

Input and Output Sample input example # #:

Sample # # of output:

1 /*This problem is very good to do, directly set SLF optimization template, run two times to take the minimum value, I began to put SLF and lll optimization together added, but because the LLL is constant card burst, card to exponential level, and then how to optimize will t off the second point, If using LLL optimization will t lose 4 points (I personally experiment), with SLF can drop, and soon. Go directly to the code. */2 3#include <algorithm>4#include <iostream>5#include <cstdlib>6#include <cstring>7#include <climits>8#include <cstdio>9#include <string>Ten#include <cmath> One#include <stack> A#include <queue> -#include <deque> - using namespacestd; the Const intgg=450000; - Const intinf=1e9; - intHead[gg]; - structnode + { -     intNext; +     intW; A     intto ; at } A[gg]; - intDis[gg]; - BOOLVis[gg]; - intc,p,pb,pa1,pa2,cnt; - intans; - intAns2; in intSum,tot; - intrans; toInlinevoidAddintIintJintW) + { -a[++cnt].to=J; thea[cnt].next=Head[i]; *a[cnt].w=W; $head[i]=CNT;Panax Notoginseng } -InlinevoidSPFA (ints) the { +deque<int>Q; Amemset (Vis,false,sizeof(Vis)); thememset (DIS,0x7f,sizeof(DIS)); +dis[s]=0; -vis[s]=true; $ Q.push_back (s); $      while(!q.empty ()) -     { -         intu=Q.front (); the Q.pop_front (); -vis[u]=false;Wuyi          for(RegisterintI=head[u]; I I=a[i].next) the         { -             intv=a[i].to; Wu             if(dis[v]>dis[u]+A[I].W) -             { Aboutdis[v]=dis[u]+A[I].W; $                 if(!Vis[v]) -                 { -vis[v]=true; -                     if(Q.empty () | | Dis[v]>Dis[q.front ()]) A                     { + Q.push_back (v); the                     } -                     Else $ Q.push_front (v); the                 } the             } the         } the     } - } in intMain () the { thescanf"%d%d%d%d%d",&c,&p,&pb,&pa1,&PA2); About      for(RegisterintI=1; i<=c; i++) the     { the         intx, y, z thescanf"%d%d%d",&x,&y,&z); + Add (x, y, z); - Add (y,x,z); the     }Bayi SPFA (PA1); theans+=dis[pa2]+DIS[PB]; the SPFA (PA2); -ans2+=dis[pa1]+DIS[PB]; -rans=min (ans,ans2); theprintf"%d\n", rans); the     return 0; the}

P3003 [Usaco10dec] apple delivery apples Delivery