Codeforces 580D 状压DP

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.

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.

题意：有n道菜，每道菜有一个愉悦值ai，你需要从中选择吃m道，有k个特殊的规则。
对于每个规则，包含x, y, c三个属性，分别代表吃完x后不吃别的菜，直接吃y能够额外获得c的愉悦值。
问你能够收获的最大的愉悦值是多少。

题解：裸的状压dp，总状态数是1<<n，for过去，dp当前状态能扩展到的状态。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;ll vis[25][25],a[25],dp[(1<<18)+5][20];int main(){ll n,m,k,i,j,x,y,z,ls;scanf("%lld%lld%lld",&n,&m,&k);for(i=1;i<=n;i++)scanf("%lld",&a[i]);for(i=1;i<=k;i++){scanf("%lld%lld%lld",&x,&y,&z);vis[x][y]=z;}ll tot=1<<n,ans=0;for(i=1;i<=n;i++)dp[1<<(i-1)][i]=a[i];for(i=0;i<tot;i++){//状态数ll cnt=0;for(ls=1;ls<=n;ls++){//这个状态最后吃的ls这道菜if((i&(1<<(ls-1)))==0)continue;++cnt;for(j=1;j<=n;j++){//接下来吃第j道菜if((i&(1<<(j-1))))continue;ll ds=i|(1<<(j-1));dp[ds][j]=max(dp[ds][j],dp[i][ls]+a[j]+vis[ls][j]);}}if(cnt==m){for(j=1;j<=n;j++){if(i&(1<<(j-1))){ans=max(ans,dp[i][j]);}}}}cout<<ans<<endl;return 0;}