# CF

1.题目描述：

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, nm and k (1 ≤ m ≤ n ≤ 180 ≤ 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 xiyi and ci (1 ≤ xi, yi ≤ n0 ≤ 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 indexes iand 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.

Examples
input
`2 2 11 12 1 1`
output
`3`
input
`4 3 21 2 3 42 1 53 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.

2.题意概述：

3.解题思路：

4.AC代码：

`#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 100100#define lson root << 1#define rson root << 1 | 1#define lent (t[root].r - t[root].l + 1)#define lenl (t[lson].r - t[lson].l + 1)#define lenr (t[rson].r - t[rson].l + 1)#define N 1111#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;int a[19], mp[20][20];ll dp[1 << 19][20];int main(){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();#endifint n, m, k;while (~scanf("%d%d%d", &n, &m, &k)){for (int i = 0; i < n; i++)scanf("%d", &a[i]);for (int i = 0; i < k; i++){int a, b, c;scanf("%d%d%d", &a, &b, &c);mp[a - 1][b - 1] = c;}memset(dp, 0, sizeof(dp));for (int i = 0; i < n; i++)dp[1 << i][i] = a[i];int tot = 1 << n;ll ans = 0;for (int s = 0; s < tot; s++) //枚举state{int cnt = 0;for (int i = 0; i < n; i++) // 枚举先吃i再吃j{if ((s & 1 << i) == 0)continue;cnt++;for (int j = 0; j < n; j++){if (s & 1 << j) // 在i之前已经吃了continue;int state = s | 1 << j;dp[state][j] = max(dp[state][j], dp[s][i] + mp[i][j] + a[j]);}}if (cnt == m)//吃够了m盘菜for (int i = 0; i < n; i++)if (s & 1 << i)ans = max(ans, dp[s][i]);}printf("%lld\n", ans);}#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);#endifreturn 0;}`

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