题意与分析（CodeForces 580D）

代码

#include <bits/stdc++.h>
#define MP make_pair
#define PB emplace_back
#define fi first
#define se second
#define ZERO(x) memset((x), 0, sizeof(x))
#define ALL(x) (x).begin(),(x).end()
#define rep(i, a, b) for (repType i = (a); i <= (b); ++i)
#define per(i, a, b) for (repType i = (a); i >= (b); --i)
#define MS(x,y) memset(x,y,sizeof(x)) 
#define QUICKIO                  \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
using namespace std;
using ll=long long;
using repType=ll;

ll dp[25][1000005];
ll taste[25];
ll extra[25][25];
int n,m,k;
ll solve(int last, int stat) // stat after eating up last
{
    //cout<<last<<" "<<stat<<endl;
    if(dp[last][stat]!=-1) return dp[last][stat];
    else
    {
        ll ans=0;
        int val=(stat-(1<<(last-1)));
        rep(i,1,n)
        {
            if(val&(1<<(i-1)))
            {
                ans=max(ans,solve(i,val)+extra[i][last]+taste[last]);
            }
        }
        //cout<<"    "<<last<<": "<<stat<<" "<<ans<<endl;
        return dp[last][stat]=ans;
    }
}
int main()
{
    MS(dp,-1);
    cin>>n>>m>>k;
    rep(i,1,n)
    {
        cin>>taste[i];
        dp[i][1<<(i-1)]=taste[i];
    }
    /*
    rep(i,1,n)
    {
        rep(j,0,((1<<n)-1))
            cout<<dp[i][j]<<" ";
        cout<<endl;
    }
    */
    rep(i,1,k)
    {
        int x,y,c;
        cin>>x>>y>>c;
        extra[x][y]=c;
    }
    ll ans=0;
    rep(i,1,n) dp[i][(1<<n)-1]=solve(i,(1<<n)-1);
    rep(i,1,n)
    {
        rep(j,0,((1<<n)-1))
        {
            if(__builtin_popcount(j)==m)
            {
                ans=max(ans,dp[i][j]);
            }
            //cout<<dp[i][j]<<" ";
        }
        //cout<<endl;
    }
    cout<<ans<<endl;
}