codeforces 580 D. Kefa and Dishes (状压dp)

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 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.

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

裸状压dp

#include 
#include 
#include 
using namespace std;
#define LL __int64

const int N = 20;

LL a[N];
LL G[N][N];
LL dp[1 << N][N];
LL num[1 << N];

void gao(int n, int m, int k) {
	memset(dp, -1, sizeof(dp));
	LL ans = 0;
	for(int i = 0; i < n; ++i) {
		dp[1 << i][i] = a[i];
		num[1 << i] = 1;
		if(m == 1) {
			ans = max(ans, a[i]);
		}
	}
	
	int maxx = (1 << n);
	for(int i = 0; i < maxx; ++i) {
		for(int j = 0; j < n; ++j) {
			if(dp[i][j] == -1) continue;
			for(int k = 0; k < n; ++k) {
				if(i & (1 << k)) continue;
				dp[i | (1 << k)][k] = max(dp[i | (1 << k)][k], dp[i][j] + a[k] + G[j][k]);
				num[i | (1 << k)] = num[i] + 1;
				if(num[i | (1 << k)] == m) {
					ans = max(ans, dp[i | (1 << k)][k]);
				}
			}
		}
	}
	cout<>n>>m>>k;
	for(int i = 0; i < n; ++i) {
		cin>>a[i];
	}
	for(int i = 0; i < k; ++i) {
		int u, v;
		LL c;
		cin>>u>>v>>c;
		--u; --v;
		G[u][v] = c;
	}
	gao(n, m, k);
	return 0;
}