版本 5e5cf9f1c0030895ce3ff346f68cb9b5aab3bf81

acm/course/Shortest_Path

Changes from 5e5cf9f1c0030895ce3ff346f68cb9b5aab3bf81 to f37e1e864b82eb2d03722188b16219027e275ccc

Shortest Path
========
Single Source Shortest Path
--------
- <big>若使用Backtracking演算法，將枚舉所有路徑，必會TLE，因此我們將列舉幾項好用DER演算法!!</big>

![](/acm/13125075_944400135668382_1681010432_n.gif)


###Relaxation
- <big>若點v離source的距離為8、點u離source的距離為3，現在有一個點k，(u,k)長度為2、(k,v)長度為1，則對(u,v)做Relax。Relax後點v離source距離更改為6。</big>

![](/acm/Relax.png)

~~~{.c }
Relex(u,v,w){
   if(dis[u]+w(u,v)<dis[v])
       dis[v]=dis[u]+w(u,v);
}
~~~

###Bellman Ford
- <big>對所有的邊做n-1次的Relax</big>
- 時間複雜度：O(VE)

![](/acm/bellman.gif)

~~~{.c }
//Pseudo Code
BellmanFord(){
    //init
    dis[source]=0;
    for(all i != source) 
        dis[i] = INF;
    //n-1 times
    for(i=0;i<n-1;i++)
        for(each edge w(u,v) in G)
            Relax(u,v,w);
}
~~~

- <big>若有負的cycle，每做一次Relax，source至cycle上點的距離，就會持續變小。</big>
    - 偵測辦法：若在第n次做Relax時，若有任何一邊可以被Relax，則有負環存在。

~~~{.c }
//Pseudo Code
BellmanFord(){
    //init
    dis[source]=0;
    for(all i != source) 
        dis[i] = INF;
    //n-1 times
    for(i=0;i<n-1;i++)
        for(each edge w(u,v) in G)
            Relax(u,v,w);
    //Negative Cycle
    for(each edge w(u,v) in G)
        if(dis[u]+w(u,v)<dis[v])
            return true;
    return false;
}
~~~

###SPFA()
- <big>相較於BellmanFord做了n-1次Relax，SPFA只對值有改變的點相連的邊做Relax。</big>
    - 方法：用Queue存有改變的點。
- <big>時間複雜度：O(kE)
- <big>時間複雜度：O(kE)</big>

~~~{.c }
SPFA(){
    for(all i){
        dis[i] = INF;
        inqueue[i] = false;
    }
    dis[source]=0;
    inqueue[source]=true;
    queue.push(source);
    while(!queue.empty()){
        int now = queue.front();
        inqueue[now]=false;
        queue.pop();
        for(each node v adjacent to now){
            if(dis[now]+w(now,v)<dis[v]){
                dis[v]=dis[now]+w(now,v);
                if(!inqueue[v]){
                    queue.push(v);
                    inqueue[v]=true;
                }
            }
        }
    }
}
~~~

All Pair Shortest Path
---------
###Floyd