CodeForces 580D.Kefa and Dishes(状压DP）

D. Kefa and Dishes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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：有n种菜，选m种。每道菜有一个权值，有些两个菜按顺序挨在一起会有combo的权
值加成。求最大权值。

题解：dp[i][j]把i转化成二进制，第几位为1表示第几盘菜被点了，j表示当前状态下最后一盘菜，dp[i][j]表示最大满意度。转移就是dp[i|(1<<⁢k)][k]=max(dp[i|(1<<⁢k)][k],dp[i][j]+ma[j][k]+s[k]);

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>typedef long long LL;using namespace std;//用二进制表示状态的话，最大也只能表示20位左右的状态，再多就基本一定会超时了int n,m,k;int sa[20];int ad[20][20];LL dp[(1<<18)+5][20];int main(){      cin>>n>>m>>k;      for(int i=0; i<n; ++i)            cin>>sa[i];      int u,v,d;      while(k--)      {            cin>>u>>v>>d;            ad[u-1][v-1]=d;      }      memset(dp,0,sizeof dp);      for(int i=0; i<n; ++i)      {            dp[1<<i][i]=sa[i];      }      long long ans=0;      int tot=1<<n;      for(int s=0; s<tot; ++s) // 枚举总状态数      {            int cnt=0;            for(int i=0; i<n; ++i) // 在已选定的菜中，选一道菜i出来在j前面吃            {                  if((s&(1<<i))==0)                        continue;                  ++cnt;                  for(int j=0; j<n; ++j) // 从未选定的菜中选一道出来，在i后面吃                  {                        if((s&(1<<j))!=0)                              continue;                        int ss=s|(1<<j);                        dp[ss][j]=max(dp[ss][j],dp[s][i]+sa[j]+ad[i][j]);                  }            }            if(cnt==m)            {                  for(int i=0; i<n; ++i)                  {                        if((s&(1<<i))!=0)                              ans=max(ans,dp[s][i]);                  }            }      }      cout<<ans<<endl;      return 0;}