最短路徑+負圈-POJ3259 蟲洞問題-圖算法基礎專項（2）

bellman-ford最短路徑算法

bellman-ford 算法,是求單源點最短路徑的算法。
我們設從起點到其他點i的最短路爲d[i]，

顯然滿足下列等式
$d[i]=min \{ d[j]+cost(j,i) | e=(j,i) \in E \}$
含義是到一點的最短路肯定是，到與他相鄰的點的最短路加上邊的權值，最小的那條。

經過多次迭代上式就可以得到最短路徑。
如果圖中不存在負圈，那麼在最短路徑中一個頂點不可能經過2次。也就是說最多|V|-1條邊。最多迭代|V|-1次，每迭代一次似乎在多加1條邊的路徑一起比較。

該算法也可以判斷有無負圈，如果迭代超過|V|-1次，有負圈。

圖表示方法

之前提到過鄰接矩陣
現在介紹另一種邊集表示法
用邊集合來表示圖

struct edge{
int from;
int to;
int cost;
}

struct edge es[MAX_E];

顯然這種表示法需要遍歷所有邊纔可能知道點與點之間有無聯繫。
這種表示法可以表示點與點之間的多重邊。
這種方法非常適用於BF算法，因爲BF算法迭代式中是針對邊的。

例題POJ 3259（蟲洞問題，判斷負圈是否存在）

Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 80546 Accepted: 29846
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).

代碼

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
//bf
//cost
//圖的邊表示
struct edge{
    int from;
    int to;
    int cost;
};
const int MAX_E=2801;
edge e[MAX_E*2];
bool negtive_circle(){
    int N,M,W;
    cin >>N>>M>>W;
    int a,b,c;
    int i=0;
    for(i=0;i<M;i++){
        cin>>a>>b>>c;
        e[2*i].from=a;
        e[2*i].to=b;
        e[2*i].cost=c;
        
        e[2*i+1].from=b;
        e[2*i+1].to=a;
        e[2*i+1].cost=c;
    }
    int j=i*2;
    for(int k=0;k<W;j++,k++){
        cin>>a>>b>>c;
        e[j].from=a;
        e[j].to=b;
        e[j].cost=-c;
    }
    
    int s=e[0].from;
    int E=j;
    vector<int> d(N+1,INT_MAX);
    d[s]=0;
    for(int v=0;v<N;v++){
        for(int i=0;i<E;i++){
            edge ee=e[i];
            //這一句很關鍵
            if(d[ee.from]!=INT_MAX&&d[ee.to]>d[ee.from]+ee.cost){
                d[ee.to]=d[ee.from]+ee.cost;
                if(v==N-1) return true;
            }
        }
    }
    return false;
}
int main(){
    int f;
    cin>>f;
    while(f--){
        bool res=negtive_circle();
        if(res)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
        
    }
    
}