Rokua P3003 [Usaco10dec] apple on delivery apple Delivery

Rokua P3003 [Usaco10dec] apple on delivery apple 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, 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 sample #: Copy

Output Example # #: Replication

idea: SPFA + SLF optimization (never made before)        Difficulty: Improve the +/province-(self-think the difficulty should be a little lower)
Next, let's talk about the (being) course of the problem.
First of all, I looked at the problem  is not running on both sides of the SPFA, and then ...

#include <algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<queue>#defineM 200005#defineMAXN 0x7fffffffusing namespaceStd;queue<int>Q;intm, n, S, t1, T2;inttot, Minn;intDis[m], vis[m];intto[m*2], net[m*2], head[m*2], cap[m*2];voidAddintUintVintW) {to[++tot] = v; Net[tot] = Head[u]; Head[u] = tot; Cap[tot] =W; to[++tot] = u; Net[tot] = Head[v]; HEAD[V] = tot; Cap[tot] =W;}voidSPFA (intx) { for(inti =1; I <= N; i++) Vis[i] =0, dis[i] =MAXN; DIS[X]=0; VIS[X] =1; Q.push (x);  while(!Q.empty ()) {        inty = Q.front (); Q.pop (); Vis[y] =0;  for(inti = head[y]; I i =Net[i]) {            intt =To[i]; if(Dis[t] > dis[y]+Cap[i]) {Dis[t]= dis[y]+Cap[i]; if(!vis[t]) vis[t] =1, Q.push (t); }        }    }}intMain () {scanf ("%d%d%d%d%d", &m, &n, &s, &AMP;T1, &T2);  for(inti =1; I <= m; i++) {        intA, B, C; scanf ("%d%d%d", &a, &b, &c);    Add (A, B, c);    } SPFA (t1); Minn= Dis[s] +DIS[T2];    SPFA (T2); Minn= MIN (Minn, dis[s]+Dis[t1]); printf ("%d", Minn); return 0;}

naked Spfa,re, two points.

It's re! I should be driving big enough, then the array opened a 0, all open long long,and then ...

#include <algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<queue>#defineLL Long Long#defineM 2000005#defineMAXN 0x7fffffffusing namespaceStd;queue<LL>Q; LL m, n, S, T1, T2; LL tot, Minn; LL Dis[m], vis[m]; LL to[m*2], net[m*2], head[m*2], cap[m*2];voidAdd (ll u, ll V, ll W) {to[++tot] = v; Net[tot] = Head[u]; Head[u] = tot; Cap[tot] =W; to[++tot] = u; Net[tot] = Head[v]; HEAD[V] = tot; Cap[tot] =W;}voidSPFA (LL x) { for(LL i =1; I <= N; i++) Vis[i] =0, dis[i] =MAXN; DIS[X]=0; VIS[X] =1; Q.push (x);  while(!Q.empty ()) {        inty = Q.front (); Q.pop (); Vis[y] =0;  for(LL i = head[y]; i; i =Net[i]) {LL T=To[i]; if(Dis[t] > dis[y]+Cap[i]) {Dis[t]= dis[y]+Cap[i]; if(!vis[t]) vis[t] =1, Q.push (t); }        }    }}intMain () {scanf ("%lld%lld%lld%lld%lld", &m, &n, &s, &AMP;T1, &T2);  for(LL i =1; I <= m; i++) {LL A, B, C; scanf ("%lld%lld%lld", &a, &b, &c);    Add (A, B, c);    } SPFA (t1); Minn= Dis[s] +DIS[T2];    SPFA (T2); Minn= MIN (Minn, dis[s]+Dis[t1]); printf ("%lld", Minn); return 0;} 

as a result, there is no egg, continue re two points.

Then I was wise (really no way) asked the elder sister, the result she told me: "This problem I remember to use SLF optimization" Qaq

But I won't.

Then learned elder sister to add the SLF after the change, but do not know why, I can not go through the example (crying)

By comparison found that the long long did not open the result is still on the drip.

#include <algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<deque>#defineM 200005#defineMAXN 0x7fffffffusing namespaceStd;deque<int>Q;intm, n, S, t1, T2;inttot, Minn;intDis[m], vis[m];intto[m*2], net[m*2], head[m*2], cap[m*2];voidAddintUintVintW) {to[++tot] = v; Net[tot] = Head[u]; Head[u] = tot; Cap[tot] =W; to[++tot] = u; Net[tot] = Head[v]; HEAD[V] = tot; Cap[tot] =W;}voidSPFA (intx) { for(inti =1; I <= N; i++) Vis[i] =0, dis[i] =MAXN; DIS[X]=0; VIS[X] =1; q.push_back (x);  while(!Q.empty ()) {        inty = Q.front (); Q.pop_front (); Vis[y] =0;  for(inti = head[y]; I i =Net[i]) {            intt =To[i]; if(Dis[t] > dis[y]+Cap[i]) {Dis[t]= dis[y]+Cap[i]; if(!Vis[t]) {Vis[t]=1; if(Q.empty () | | dis[t]<Dis[q.front ()]) Q.push_front (t); ElseQ.push_back (t); }            }        }    }}intMain () {scanf ("%d%d%d%d%d", &m, &n, &s, &AMP;T1, &T2);  for(inti =1; I <= m; i++) {        intA, B, C; scanf ("%d%d%d", &a, &b, &c);    Add (A, B, c);    } SPFA (t1); Minn= Dis[s] +DIS[T2];    SPFA (T2); Minn= MIN (Minn, dis[s]+Dis[t1]); printf ("%d", Minn); return 0;}

long long changed back to int and then A.

10,000 heads in my heart * * * * *

But fortunately, finally, it's a.

Rokua P3003 [Usaco10dec] apple on delivery apple Delivery