D - Kefa and Dishes
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 krules 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 indexes i 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.
Examples
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：给了n个菜，吃菜能获得满意度，还有一些规则就是在吃y菜之前吃x菜能额外获得c点满意度，问吃m个菜能获得的最大满意度是多少。
思路：状压dp裸题，从n的大小就可以知道，dp[i][j] 表示以吃菜的集合为i（二进制1已吃，0未吃），j为最后吃的菜，
转移方程：dp[i|(1<<(k-1)][k] = max(dp[i|(1<<(k-1)][k], dp[i][j] + c[k] + ad[j][k])
#include <bits/stdc++.h>
//#include "dbg_.h"
using namespace std;
const int N = 30;
int ad[20][20];
long long c[20];
long long dp[1<<18][19];
int main() {
int n, m, k;
scanf ("%d %d %d", &n, &m, &k);
for (int i = 1; i <= n; ++i) scanf ("%lld", &c[i]);
int u, v;
long long w;
for (int i = 1; i <= k; ++i) {
scanf ("%d %d %lld", &u, &v, &w);
ad[u][v] = w;
}
for (int i = 1; i <= n; ++i) dp[1<<(i-1)][i] = c[i];
long long ans = 0;
for (int i = 1; i < (1<<n); ++i) {
for (int j = 1; j <= n; ++j) {
if ((i&(1<<(j-1)))) { //找一个以到达的状态
for (int k = 1; k <= n; ++k) {
if (!(i&(1<<(k-1)))) { //找未到达状态，从以到达的状态转移到未到达的状态
dp[i|(1<<(k-1))][k] = max(dp[i|(1<<(k-1))][k], dp[i][j] + c[k] + ad[j][k]);
//printf ("ab: %d %d %lld\n", k, j, dp[i|(1<<(j-1))][j]);
}
}
}
}
}
for (int i = 1; i < (1 << n); ++i) {
if (__builtin_popcount(i) == m) {
for (int k = 1; k <= n; ++k) {
ans = max(ans, dp[i][k]);
}
}
}
printf ("%lld\n", ans);
return 0;
}