Codeforces Round #321 (Div. 2) D 狀壓dp

連結：戳這裡

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.

題意：

給出n道菜，要點m道菜，每道菜有一個滿意值ai,給出K個規則，表示u,v兩道菜連著點的話滿意度+c[u][v]

問怎麼點這m道菜使得滿意度最大、

思路：

一看就是狀壓dp，設定dp[i][j] i表示已經選了的菜 i&(1<<j) j表示當前菜中，j作為最後一道點的菜

那麼我們從已經點了的菜裡面選擇一道菜，再去選當前的第j道菜組成c[i][j] 每次都n*n列舉

那麼轉移方程就是 dp[now][j]=max(dp[now][j],dp[i][j]+c[j][k]+a[k])

嗯 我第一次寫 不是很明白

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int c[110][110];
ll dp[1000100][20];
int a[20];
int n,m,K;
int main(){
    scanf("%d%d%d",&n,&m,&K);
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=K;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        u--;v--;
        c[u][v]=w;
    }
    for(int i=0;i<n;i++) dp[1<<i][i]=a[i];
    ll ans=0;
    for(int i=0;i<(1<<n);i++){
        int cnt=0;
        for(int j=0;j<n;j++){
            if((i&(1<<j))==0) continue; /// 選了第j道菜
            cnt++;
            for(int k=0;k<n;k++){
                if((i&(1<<k))>0) continue; /// 第k道菜作為當前最後選的菜
                int tmp=i|(1<<k);
                dp[tmp][k]=max(dp[tmp][k],dp[i][j]+c[j][k]+a[k]);
            }
        }
        if(cnt==m) {
            for(int j=0;j<n;j++){
                if((i&(1<<j))>0) ans=max(ans,dp[i][j]);
            }
        }
    }
    printf("%I64d\n",ans);
    return 0;
}