bellman-ford 算法 spfa 算法 实例 poj 3259

#include<iostream>#include<stdio.h>using namespace std;#define max 100000struct side{       int start;       int end;       int value;};side a[5000];int ver[500];int m1,m2,m3;bool bellman(){     for(int i=1;i<=m1;i++){                ver[i]=max;                }     ver[1]=0;     int flag;     int h=0;     for(;h<m1-1;h++){             int g=1;             flag=0;             for(;g<=m2+m3;g++){                     if(ver[a[g].start]+a[g].value<ver[a[g].end]){                              ver[a[g].end]=ver[a[g].start]+a[g].value;                              flag=1;                                   }                     if(ver[a[g].end]+a[g].value<ver[a[g].start]&&a[g].value>=0){                              ver[a[g].start]=ver[a[g].end]+a[g].value;                                 flag=1;                              }                                }             if(flag==0)                  return true;                  }     if(h>=m1-1&&flag==1)             return false;             }             int main(){       int n;    cin>>n;    while(n--){         scanf("%d%d%d",&m1,&m2,&m3);        //cin>>m1>>m2>>m3;                int temp1,temp2,temp3;        int j=1;        for(;j<=m2;j++){               // cin>>a[j].start>>a[j].end>>a[j].value;               scanf("%d%d%d",&temp1,&temp2,&temp3);               a[j].start=temp1;               a[j].end=temp2;               a[j].value=temp3;                }        for( ;j<=m2+m3;j++){                 scanf("%d%d%d",&temp1,&temp2,&temp3);                //cin>>a[j].start;                //cin>>a[j].end;                a[j].start=temp1;                a[j].end=temp2;           //     int temp;             //   cin>>temp;                a[j].value=-temp3;                }        if(m1==1){                  printf("YES\n");            //      cout<<"YES";         //         if(n>0)           //           printf("\n");                  //    cout<<endl;                  continue;                  }        if(bellman())               printf("NO\n");         //     cout<<"NO";       else                     // cout<<"YES";             printf("YES\n");      /*  if(n>0)               cout<<endl;               }*/               }        return 0;        }                

 

bellman-ford 是对dijkstra算法的优化。其主要的特点就是能够判断图中是否存在负环。

负环即一个环路中的节点间的距离和为负数。

如果存在负环，那么其始点到各点的最短路径和会不断地更新，即该环路中不存在最短距离。

bellman-ford 的基本思想：

            1. 用struct结构体来表示边；其内容有：起点，终点，边权。

            2. 所有节点的权值初始化为无穷大，始点赋为0。

            3. 由边来更新节点的权值；如果始点+边<终点，则更新终点，有多少条边就做多少次。

            4. 将第三步循环做，直到不存在更新（跳出，且图的始点到各点的最短路径得出）或循环满了n-1次后，还存在更新（图中存在负环）。

            5. 判断是否存在负环。

 

 

spfa 即对bellman-ford 得一个优化；主要是针对于在更新节点时，不需要将每一个节点都遍历一次。

一般spfa 都用邻接表的方法来存储图。

spfa 主要思想:

             1. 从原点开始，遍历每一个与之相连的节点，将每一个更新的且不在队列中的节点入队，以便下一次的更新。

             2. 提取队首的节点，重复1，知道队列为空。

             3. 如果当存在节点第n次入队时，则可判定该图存在负环。

 

 

A - Wormholes

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

 

             

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