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