CF 580D Kefa and Dishes(简单状压dp）

题目链接

When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There weren dishes. Kefa knows that he needs exactlym dishes. But at that, he doesn't want to order the same dish twice to taste as many dishes as possible.

Kefa knows that the i-th dish gives hima_i units of satisfaction. But some dishes do not go well together and some dishes go very well together. Kefa set to himself k rules of eating food of the following type — if he eats dishx exactly before dishy (there should be no other dishes betweenx andy), then his satisfaction level raises byc.

Of course, our parrot wants to get some maximal possible satisfaction from going to the restaurant. Help him in this hard task!

Input

The first line of the input contains three space-separated numbers,n,m andk (1 ≤ m ≤ n ≤ 18,0 ≤ k ≤ n * (n - 1)) — the number of dishes on the menu, the number of portions Kefa needs to eat to get full and the number of eating rules.

The second line contains n space-separated numbersa_i, (0 ≤ a_i ≤ 10⁹) — the satisfaction he gets from the i-th dish.

Next k lines contain the rules. Thei-th rule is described by the three numbersx_i,y_i andc_i (1 ≤ x_i, y_i ≤ n,0 ≤ c_i ≤ 10⁹). That means that if you eat dish x_i right before dishy_i, then the Kefa's satisfaction increases byc_i. It is guaranteed that there are no such pairs of indexesi andj (1 ≤ i < j ≤ k), that x_i = x_j and y_i = y_j.

Output

In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.

Sample test(s)

Input

2 2 1
1 1
2 1 1

Output

3

Input

4 3 2
1 2 3 4
2 1 5
3 4 2

Output

12

Note

In the first sample it is best to first eat the second dish, then the first one. Then we get one unit of satisfaction for each dish and plus one more for the rule.

In the second test the fitting sequences of choice are 4 2 1 or 2 1 4. In both cases we get satisfaction 7 for dishes and also, if we fulfill rule 1, we get an additional satisfaction 5.

题意：

第一行n,m,k代表有n种菜，每种菜都有一定的满意度，让你从中选择m种吃。还有些菜你按照他给的顺序吃满意度还会增加，有k种；

第二行为n种菜的满意度；

接下来k行x,y,c代表先吃x再吃y满意度增加c。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL __int64
using namespace std;
LL dp[270000][20];
LL a[20],b[20][20];
int get(int x)
{
    int ans=0;
    for(;x;x-=x&(-x)) ans++;
    return ans;
}
int main()
{
    int n,m,K;
    while(scanf("%d%d%d",&n,&m,&K)!=EOF)
    {
        for(int i=0;i<n;i++) scanf("%I64d",&a[i]);
        memset(b,0,sizeof(b));
        for(int i=1;i<=K;i++)
        {
            int x,y,c;
            scanf("%d%d%d",&x,&y,&c);
            b[--x][--y]=c;///按照顺序吃增加的满意度
        }
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++) dp[(1<<i)][i]=a[i];
        LL ans=0;
        for(int i=0;i<(1<<n);i++)
        {
            for(int j=0;j<n;j++)
            {
                if(!(i&(1<<j))) continue;
                for(int k=0;k<n;k++)
                {
                    if(j==k) continue;
                    if(!(i&(1<<k))) continue;
                    dp[i][j]=max(dp[i][j],dp[i^(1<<j)][k]+a[j]+b[k][j]);
                }
            }
            if(get(i)==m)///选择了m种，更新答案
            {
                for(int j=0;j<n;j++) ans=max(ans,dp[i][j]);
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}