POJ 3268 Silver Cow Party SPFA

POJ 3268 Silver Cow party SPFA.

Free to have nothing to do with water a cow problem sleep well. It's going to be 11 right now. "Tears

Description:n (1 <= n <=), m stripe weight (cost) has a forward edge (1 <= m <= 1e5), given a target point T, to find the value of the maximum value.

We define the value of each point. The value of a point = the minimum cost of the current point to the target point and the minimum cost of the target point to the current point.

Solution:

Well.. The solution is simple. Run SPFA on T and turn the side upside down SPFA.

Why do you do this? " Although obviously. But in order to get the logical level of the problem analysis, write it. 】

We consider the value of each point. Let's consider the minimum cost of the target point to the current point. This is easy to ask. SPFA to T run once.

Value of point = Minimum cost of the target point to the current point * 2?

Wrong. There is a direction graph. Unidirectional.

The simple SPFA to find the answer, a legal side is so defined, you can reach V, we can expand.

The reason we get the answer is that you can reach V when V does not necessarily reach U. Let's change the condition of the SPFA extension node: if V can reach U, it is considered as having an edge between U and V.

So we SPFA the side reversal. Then find out the maximum value of reverse_dis[i] + dis[i] just fine.

Slag Code:

1#include <iostream>2#include <string.h>3#include <stdio.h>4#include <algorithm>5#include <math.h>6#include <stdlib.h>7 using namespacestd;8 Const intM = 1e5 +5;9 Const intN = ++5;Ten structqwq{intV, Val, next;} EDGE[M], revedge[m]; One BOOLVis[n]; A intCNT =0, revcnt =0, N, M, T; - intQ[n *N], head[n], revhead[n], dis[n], revdis[n];  - voidAddintUuintvvintTT) the { -edge[++ cnt] = (QWQ) {VV, TT, Head[uu]}; HEAD[UU] =CNT; -revedge[++ revcnt] = (QWQ) {UU, TT, REVHEAD[VV]}; REVHEAD[VV] =revcnt; - } + voidsolve () { -memset (Vis,0,sizeof(Vis)); +memset (Q,0,sizeof(q)); Amemset (DIS,0x3f,sizeof(DIS)); at     //Back route, normal -VIS[T] =1; -     inth =0, t =0; -q[++ T] =T; -DIS[T] =0; -      while(H <t) { in         intUU = q[++ h]; VIS[UU] =0; -          for(inti = Head[uu]; ~i; i =Edge[i].next) { to             intv =edge[i].v; +             if(Dis[v] > Dis[uu] + edge[i].val) {//Relax First -DIS[V] = Dis[uu] +Edge[i].val; the                 if(!Vis[v]) { *VIS[V] =1; q[++ T] =v; $                 }Panax Notoginseng             } -         } the     } +memset (Vis,0,sizeof(Vis)); Amemset (Q,0,sizeof(q)); thememset (Revdis,0x3f,sizeof(Revdis)); +     //go route, reverse mode Qaq -VIS[T] =1; $h =0, t =0; $q[++ T] =T; -REVDIS[T] =0; -      while(H <t) { the         intUU = q[++ h]; VIS[UU] =0; -          for(inti = Revhead[uu]; ~i; i =Revedge[i].next) {Wuyi             intv =revedge[i].v; the             if(Revdis[v] > Revdis[uu] + revedge[i].val) {//Relax Qaq First . -REVDIS[V] = Revdis[uu] +Revedge[i].val; Wu                 if(!Vis[v]) { -VIS[V] =1; q[++ T] =v; About                 } $             } -         } -     } -     intMX =0; A      for(inti =1; I <= N; i + +){ +         //printf ("%d%d\n", Dis[i], revdis[i]); the         if(I! = T && dis[i] + revdis[i] > mx) mx = dis[i] +Revdis[i]; -     } $printf"%d\n", MX); the } the intMain () the {     theMemset (Head,-1,sizeof(head)); -memset (Revhead,-1,sizeof(Revhead)); inscanf"%d%d%d", &n, &m, &T); the      for(inti =1; I <= m; i + +){ the         intUU, VV, TT; Aboutscanf"%d%d%d", &uu, &AMP;VV, &TT); the Add (UU, VV, TT); the     } the solve (); +     return 0; -}

POJ 3268 Silver Cow Party

POJ 3268 Silver Cow Party SPFA