---- shortest path problem in graph theory

Shortest Path Problem There are three common methods:

Floyd,Dijkstra,SPFA

The following is a summary of the code (reference)

 

Floyd

FIG rectifiable shortest between any two

A great shortage ---- O (N ^ 3) time complexity

Code difficulty simple

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
ll n,m,k;
ll s[1005][1005];
int main()
{
    memset(s,0x3f,sizeof(s));
    scanf("%lld%lld%lld",&n,&m,&k);
    for(ll i=1;i<=m;i++)
    {
        ll x,y,z;
        scanf("%lld%lld%lld",&x,&y,&z);
        s [x] [y] = min (p [x] [y], z);
    }
    for(ll i=1;i<=n;i++)
    {
        s[i][i]=0;
        for(ll j=1;j<=n;j++)
        {
            s[j][j]=0;
            for(ll o=1;o<=n;o++)
            {
                s[o][o]=0;
                s [j] [i] = min (s [j] [i], s [j] [i] + s [i] [on]);
            }
        }
    }
    for(ll i=1;i<=k;i++)
    {
        ll b,e;
        scanf("%lld%lld",&b,&e);
        if(s[b][e]<=0x3f3f3f3f/2)
        printf("%lld\n",s[b][e]);
        else printf("impossible\n");
    }
    return 0;
}

Floyd Code

 

Dijkstra

Rectifiable single-source shortest

Usually time complexity ---- O (n ^ 2)

Optimization of the time complexity ---- O (m logn)

Code general difficulty

 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 ll n,m,now;
 5 ll s[1005][1005],dis[1005];
 6 bool vi[1005];
 7 int main()
 8 {
 9     scanf("%lld%lld",&n,&m);
10     memset(s,0x3f,sizeof(s));
11     for(ll i=1;i<=n;i++) s[i][i]=0;
12     for(ll i=1;i<=m;i++)
13     {
14         ll x,y,z;
15         scanf("%lld%lld%lld",&x,&y,&z);
16         s[x][y]=min(s[x][y],z);
17     }
18     memset(dis,0x3f,sizeof(dis));
19     dis[1]=0;
20     for(ll i=1;i<n;i++)
21     {
22         ll t=0;
23         for(ll j=1;j<=n;j++)
24         {
25             if((dis[j]<dis[t]||t==0)&&!vi[j])
26             {
27                 t=j;
28             }
29         }
30         vi[t]=1;
31         for(ll j=1;j<=n;j++)
32         {
33             dis[j]=min(dis[j],dis[t]+s[t][j]);
34         }
35     }
36     if(dis[n]<0x3f3f3f3f/2)
37     printf("%lld",dis[n]);
38     else printf("-1");
39     return 0;
40 }

General Dj Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef pair<long long,long long> pp;
 4 typedef long long ll;
 5 ll n,m;
 6 ll fir[100005],to[100005],nxt[100005],w[100005],num;
 7 ll dis[100005];
 8 bool vis[100005];
 9 void add(ll x,ll y,ll z)
10 {
11     nxt[++num]=fir[x];
12     fir[x]=num;
13     to[num]=y;
14     w[num]=z;
15 }
16 void dj()
17 {
18     memset(dis,0x3f,sizeof(dis));
19     memset(vis,0,sizeof(vis));
20     priority_queue<pp,vector<pp>,greater<pp> > q;
21     q.push(make_pair(0,1));
22     dis[1]=0;
23     while(q.size())
24     {
25         ll u=q.top().second,diss=q.top().first;
26         q.pop();
27         if(vis[u]) continue;
28         vis[u]=1;
29         for(ll i=fir[u];i;i=nxt[i])
30         {
31             ll v=to[i];
32             if(dis[v]>dis[u]+w[i])
33             {
34                 dis[v]=dis[u]+w[i];
35                 
36                 q.push(make_pair(dis[v],v));
37             }
38         }
39     }
40 }
41 int main()
42 {
43     scanf("%lld%lld",&n,&m);
44     for(ll i=1;i<=m;i++)
45     {
46         ll x,y,z;
47         scanf("%lld%lld%lld",&x,&y,&z);
48         add(x,y,z);
49     }
50     dj();
51     if(dis[n]<0x3f3f3f3f/2) printf("%lld",dis[n]);
52     else printf("-1");
53     return 0;
54 }

Dj Code Optimization

 

SPFA

Rectifiable single-source shortest

Time complexity ---- O (km)

Code general difficulty

 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 struct edge
 5 {
 6     ll nxt,to,w;
 7 }e[100005];
 8 ll n,m;
 9 ll num,fir[100005],dis[100005],vis[100005];
10 void add(ll x,ll y,ll z)
11 {
12     e[++num].nxt=fir[x];
13     fir[x]=num;
14     e[num].to=y;
15     e[num].w=z;
16 }
17 void spfa()
18 {
19     memset(dis,0x3f,sizeof(dis));
20     dis[1]=0;
21     queue<ll> q;
22     q.push(1);
23     while(q.size())
24     {
25         ll u=q.front();q.pop();
26         vis[u]=0;
27         for(ll i=fir[u],v;v=e[i].to,i;i=e[i].nxt)
28         {
29             if(dis[v]>dis[u]+e[i].w)
30             {
31                 dis[v]=dis[u]+e[i].w;
32                 if(!vis[v])
33                 {
34                     vis[v]=1;
35                     q.push(v);
36                 }
37             }
38         }
39     }
40 }
41 int main()
42 {
43     scanf("%lld%lld",&n,&m);
44     for(ll i=1;i<=m;i++)
45     {
46         ll x,y,z;
47         scanf("%lld%lld%lld",&x,&y,&z);
48         add(x,y,z);
49     }
50     spfa();
51     if(dis[n]<0x3f3f3f3f/2)
52     printf("%lld",dis[n]);
53     else printf("impossible");
54     return 0;
55 }

SPFA Code

 

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