最短路之貝爾曼（bellmanford）算法

貝爾曼（bellmanford）算法：

大意：分爲兩步：1，對各邊進行鬆弛操做，即更新最短距離 2，判斷是否產生迴路

 

Bellman-Ford算法的流程以下：
給定圖G(V, E)（其中V、E分別爲圖G的頂點集與邊集），源點s，
數組Distant[i]記錄從源點s到頂點i的路徑長度，初始化數組Distant[n]爲, Distant[s]爲0；

 如下操做循環執行至多n-1次，n爲頂點數：
對於每一條邊e(u, v)，若是Distant[u] + w(u, v)< Distant[v]，則另Distant[v] = Distant[u]+w(u, v)。w(u, v)爲邊e(u,v)的權值；
若上述操做沒有對Distant進行更新，說明最短路徑已經查找完畢，或者部分點不可達，跳出循環。不然執行下次循環；
爲了檢測圖中是否存在負環路，即權值之和小於0的環路。對於每一條邊e(u, v)，若是存在Distant[u] + w(u, v)< Distant[v]的邊，則圖中存在負環路，便是說改圖沒法求出單源最短路徑。不然數組Distant[n]中記錄的就是源點s到各頂點的最短路徑長度。
可知， Bellman-Ford算法 尋找單源最短路徑的時間複雜度爲O(V*E).

 

例子：

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

Sample Input

2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8

Sample Output

NO YES

題目大意：

　　　　FJ童鞋家的農田裏面有若干個蟲洞，有一些蟲洞是相連的，能夠消耗一些時間爬過去(無向)。還有一些蟲洞比較特殊，從一頭爬到那一頭以後時間會退後一些(有向)

　　　　FJ但願經過爬蟲洞(?)來時時間退後，從而能看到另一個本身。根據輸入的蟲洞信息來判斷是否能夠實現FJ的願望。

code：

  #include<stdio.h>
#define inf 999999
struct node
{
    int u,v,w;
} edge[6000];
int dis[505];
int n,m,w,in;
void add(int u,int v,int c)
{
    in++;
    edge[in].u=u;
    edge[in].v=v;
    edge[in].w=c;
}
int bellman()
{
    int u,v,w,i,j,flag;
    for(i=1; i<=n; i++)//初始化
    {
        dis[i]=inf;
    }
    dis[1]=0;
    flag=0;
    for(i=1; i<=n; i++)//邊的鬆弛
    {
        for(j=1; j<=in; j++)
        {          

             if(dis[edge[j].v]>dis[edge[j].u]+edge[j].w)
            {
                dis[edge[j].v]=dis[edge[j].u]+edge[j].w;
            }
        }
    }
    for(i=1; i<=in; i++)//判斷是否有迴路
        if(dis[edge[i].v]>dis[edge[i].u]+edge[i].w)
            return 1;
    return 0;
}
int main()
{
    int i,j,t,u,v,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&w);
        in=0;
        for(i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            add(u,v,c);
            add(v,u,c);
        }
        for(i=0; i<w; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            add(u,v,-1*c);
        }
        if(bellman())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
Dijkstra算法和Bellman算法思想有很大的區別：
      Dijkstra算法在求解過程當中，源點到集合S內各頂點的最短路徑一旦求出，則以後不變了，修改的僅僅是源點到T集合中各頂點的最短路徑長度。Bellman算法在求解過程當中，每次循環都要修改全部頂點的dist[]，也就是說源點到各頂點最短路徑長度一直要到Bellman算法結束才肯定下來。
若是存在從源點可達的負權值迴路，則最短路徑不存在，由於能夠重複走這個迴路，使得路徑無窮小。     

 做爲一種單源最短路徑算法，Bellman-Ford對於有向圖和無向圖都能適用，它還有一個Dijkstra算法沒法具有的特色，那就是對含負權圖的最短路徑搜索。每i輪對邊的遍歷以後，只要不存在負權迴路，Bellman-Ford算法均可以保證得到距離源點i條邊的點的最短路徑。由於最短路徑的最大長度不會超過n條邊，n爲節點的數目。因此n次遍歷後全部的節點一定都能找到最短路徑。若是n次後還能夠繼續鬆弛，則代表該圖存在負權迴路，能夠對代碼稍做修改來計算負權環的大小。