Codeforces#321 (Div. 2) D. Kefa and Dishes（DP，掩码）

D. Kefa and Dishes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m 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 him ai 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 dish x exactly before dish y (there should be no other dishes between x and y), then his satisfaction level raises by c.

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 and k (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 numbers ai, (0 ≤ ai ≤ 109) — the satisfaction he gets from the i-th dish.

Next k lines contain the rules. The i-th rule is described by the three numbers xi, yi and ci (1 ≤ xi, yi ≤ n, 0 ≤ ci ≤ 109). That means that if you eat dish xi right before dish yi, then the Kefa's satisfaction increases by ci. It is guaranteed that there are no such pairs of indexesi and j (1 ≤ i < j ≤ k), that xi = xj and yi = yj.

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：Kefa 去吃东西，菜单上有n中菜可以选，每种菜分别会带给kefa ai的满足感，此外有k条规则，给出u，v，c，先吃u再吃v会得到c的满足感。kefa要吃m种菜，求得到满足感的最大值。

思路：用一个二维的dp数组，第一个元素用掩码表示已经拿的菜，注意大小是2^18，第二个元素表示拿的最后一种菜，先置dp数组为-1，以此来消除掩码没有意义的空元素，可以知道，初始状态是第i种菜先拿的时候，值为ai，状态转移，后面的状态有原来dp[][j] ,或者是先拿一个i再拿j得到，为dp[]+c[j]+g[i][j]。最后遍历得到拿的菜的数量为m的dp的最大值就是答案了。

#include
#include
#include
#include
using namespace std;

#define LL long long  

LL dp[500000][20];
LL c[20];
LL g[20][20];

LL getBitNum(LL x)
{
    LL ret = 0;
    for (LL i = 0; i <= 20;i++)
    if (1 << (i - 1) & x)
        ret++;
    return ret;
}

LL setBit(LL bitmasks, LL pos)
{
    return bitmasks | (1 << (pos - 1));
}

bool isin(LL bitmasks, LL pos)
{
    return bitmasks & (1 << (pos - 1));
}

int main()
{
    LL n, m, k;
    while (scanf("%I64d%I64d%I64d", &n, &m, &k) != EOF)
    {
        for (LL i = 1; i <= n; i++)
            scanf("%I64d", &c[i]);
        LL u, v, cc;
        for (LL i = 0; i < k; i++)
        {
            scanf("%I64d%I64d%I64d", &u, &v, &cc);
            g[u][v] = cc;
        }

        memset(dp, -1, sizeof(dp));
        for (LL i = 1; i <= n; i++)
        {
            dp[setBit(0, i)][i] = c[i];
        }

        LL maxbit = (1 << n) - 1;
        for (LL bit = 0; bit <= maxbit;bit++)
        for (LL i = 1; i <= n; i++)     //从i开始点
        {
            if (dp[bit][i] == -1) continue;
            for (LL j = 1; j <= n; j++)     
            if (i != j && !isin(bit, j))
            {
                dp[setBit(bit, j)][j] = max(dp[setBit(bit, j)][j], dp[bit][i] + c[j] + g[j][i]);
            }
        }
        LL ans = 0;
        for (LL i = 0; i <= maxbit;i++)
        if (getBitNum(i)==m)
        for (LL j = 1; j <= n; j++)
            ans = max(ans, dp[i][j]);
        cout << ans << endl;
    }
}