Flody算法求解多源最短路问题

蓝桥公园

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=409;
int n,m,q,d[N][N];
signed main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
  cin>>n>>m>>q;
  memset(d,0x3f,sizeof(d));//初始化d极大值
  for(int i=1;i<=n;i++)d[i][i]=0;//自己到自己距离为0
  while(m--){
    int u,v,w;cin>>u>>v>>w;
    d[u][v]=min(d[u][v],w);//双向图，可能有重边，取小
    d[v][u]=min(d[v][u],w);
  }
  for(int k=1;k<=n;k++)//Flody
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
  while(q--){//q次询问
    int st,ed;cin>>st>>ed;
    if(d[st][ed]>=0x3f3f3f3f3f3f3f3f)cout<<"-1"<<'\n';//2e18也可
    else cout<<d[st][ed]<<'\n';
  }
}

小蓝组网

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=505;
int n,m,d[N][N];
bool ok=1;
signed main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
  cin>>n>>m;
  memset(d,0x3f,sizeof(d));
  for(int i=1;i<=n;i++)d[i][i]=0;
  while(m--){
    int a,b;cin>>a>>b;
    d[a][b]=min(d[a][b],1LL);
    d[b][a]=min(d[b][a],1LL);
  }
  for(int k=1;k<=n;k++)
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
  int ans=0;
  for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++){
      if(d[i][j]>=0x3f3f3f3f3f3f3f3f){
        ok=0;
        continue;
        }
      ans=max(ans,d[i][j]);
    }
  if(!ok)cout<<"NO"<<'\n';
  else cout<<"YES"<<'\n';
  cout<<ans;
}

最短路问题

#include <bits/stdc++.h>
using namespace std;
const int N = 305;
int n,q,g[N][N];
bool is_prime(int x){
    if(x<2)return 0;
    for (int i = 2; i*i <= x ; i ++ )
        if (x % i == 0)
            return 0;
    return 1;
}
void floyd(){
    for (int k = 1; k <= n; ++ k )
        for (int i = 1; i <= n; ++ i )
            for (int j = 1; j <= n; ++ j )
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
int main(){
    cin >> n >> q; 
    memset(g, 0x3f, sizeof g);
    for (int i = 1; i <= n; ++ i )
        for (int j = i + 1; j <= n; ++ j )
            if (is_prime(i) + is_prime(j) == 1)
                g[i][j] = i*j/__gcd(i,j);
    
    floyd();
    
    while (q -- ){
        int a, b;cin >> a >> b;
        if (g[a][b] == 0x3f3f3f3f3f3f)cout<<"-1"<<'\n';
        else
            cout << g[a][b] <<'\n';
    }
}

会面

#include <bits/stdc++.h>
using namespace std;
const int N = 209;
int n,m,g[N][N];
int s1,e1,s2,e2;
void floyd(){
    for (int k = 1; k <= n; ++ k )
        for (int i = 1; i <= n; ++ i )
            for (int j = 1; j <= n; ++ j )
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
}
int main(){
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    for (int i = 1; i <= n; ++ i )g[i][i] = 0;
    while (m -- ){
        int a, b, c;cin>>a>>b>>c;
        g[a][b]=min(g[a][b],c);
        g[b][a]=min(g[b][a],c);
    }
    cin >> s1 >> e1 >> s2 >> e2;
    
    floyd();
    
    int res=0x3f3f3f3f3f;
    for (int k = 1; k <= n; ++ k )
        res=min(res,max(g[s1][k],g[s2][k])+max(g[k][e1],g[k][e2]));
    cout<<(res >= 0x3f3f3f3f3f? -1: res) << '\n';
}