• Bellman算法


    Bellman算法

    当图有负圈的时候可以用这个判断最短路!

    【时间复杂度】O((nm))

    &代码:

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int INF = 0x3f3f3f3f;
    #define cle(a,val) memset(a,(val),sizeof(a))
    #define SI(N) scanf("%d",&(N))
    #define SII(N,M) scanf("%d %d",&(N),&(M))
    #define SIII(N,M,K) scanf("%d %d %d",&(N),&(M),&(K))
    #define rep(i,b) for(int i=0;i<(b);i++)
    #define rez(i,a,b) for(int i=(a);i<=(b);i++)
    #define red(i,a,b) for(int i=(a);i>=(b);i--)
    const ll LINF = 0x3f3f3f3f3f3f3f3f;
    #define PU(x) puts(#x);
    #define PI(A) cout<<(A)<<endl;
    #define DG(x) cout<<#x<<"="<<(x)<<endl;
    #define DGG(x,y) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<endl;
    #define DGGG(x,y,z) cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<" "<<#z<<"="<<(z)<<endl;
    #define PIar(a,n) rep(i,n)cout<<a[i]<<" ";cout<<endl;
    #define PIarr(a,n,m) rep(aa,n){rep(bb, m)cout<<a[aa][bb]<<" ";cout<<endl;}
    const double EPS = 1e-9 ;
    /*  ////////////////////////   C o d i n g  S p a c e   ////////////////////////  */
    const int MAXN = 1000 + 9 ;
    struct Edge{
        int from,to,dist;
        Edge(int u,int v,int d):from(u),to(v),dist(d){}
    };
    vector<Edge> edges;
    vector<int> G[MAXN];
    void Init(int n){
        rep(i,n) G[i].clear();
        edges.clear();
    }
    void AddEdge(int u,int v,int d){
        edges.push_back(Edge(u,v,d));
        G[u].push_back(edges.size()-1);
    }
    int n,m;
    int inq[MAXN],cnt[MAXN],d[MAXN],p[MAXN];
    bool BellmanFord(int s){
        queue<int> Q;
        memset(inq,0,sizeof(inq));
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<n;i++)
            d[i]=INF;
        d[s]=0;
        inq[s]=true;
        Q.push(s);
        while(!Q.empty()){
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for (int i=0;i<G[u].size();i++){
                Edge& e=edges[G[u][i]];
                if (d[u]<INF&&d[e.to]>d[u]+e.dist){
                    d[e.to]=d[u]+e.dist;
                    p[e.to]=G[u][i];    
                    if (!inq[e.to]){
                        Q.push(e.to);
                        inq[e.to]=true;
                        if (++cnt[e.to]>n) return false;
                    }
                }
            }
        }
    }
    void Solve()
    {
        while(cin>>n>>m){
            int a,b,c;
            Init(n);
            for(int i=0;i<n;i++){
                scanf("%d%d%d",&a,&b,&c);
                //the number input is started from 1,but the code is started from 0. so must -1
                a--,b--;
                AddEdge(a,b,c);
            }
            BellmanFord(0);
            PIar(d,n)
        }
    }
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("1.in", "r", stdin);
        freopen("1.out","w",stdout);
    #endif
    //iostream::sync_with_stdio(false);
    //cin.tie(0), cout.tie(0);
        // int T;cin>>T;while(T--)
        Solve();
        return 0;
    }
    

    测试:
    输入
    5 5
    1 2 1
    1 3 7
    2 5 4
    3 4 3
    3 5 2
    输出
    0 1 7 10 5

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  • 原文地址:https://www.cnblogs.com/s1124yy/p/6031307.html
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