#include <bits/stdc++.h>

using namespace std;

#define rep(i, l, r) for (int i = l; i <= r; i++)
// #define int long long
using db = double;
using ll = long long;
using ull = unsigned long long;
const int INF = 1 << 30;
const long long INFL = 1LL << 60;

int f[124][124];

signed main()
{
    memset(f, 0x3f3f3f3f, sizeof(f));
    int a,b;
    cin>>a>>b;
    int n=3;
    f[1][2]=f[2][1]=a;
    f[2][3]=f[3][2]=b;
    rep(k, 1, n)
        rep(i, 1, n)
            rep(j, 1, n) 
                    f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
    cout<<f[1][3];
}

思路是设 $u,t,v$ 然后 $w(u,t)=a,w(t,v)=b$.然后求最短路.

https://www.luogu.com.cn/record/120564071