Codeforces Round #321 (Div. 2) D 状压dp

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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：

给出n道菜，要点m道菜，每道菜有一个满意值ai,给出K个规则，表示u,v两道菜连着点的话满意度+c[u][v]

问怎么点这m道菜使得满意度最大、

思路：

一看就是状压dp，设置dp[i][j] i表示已经选了的菜 i&(1<<j) j表示当前菜中，j作为最后一道点的菜

那么我们从已经点了的菜里面选择一道菜，再去选当前的第j道菜组成c[i][j] 每次都n*n枚举

那么转移方程就是 dp[now][j]=max(dp[now][j],dp[i][j]+c[j][k]+a[k])

嗯 我第一次写 不是很明白

代码：

#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; /// 选了第j道菜
            cnt++;
            for(int k=0;k<n;k++){
                if((i&(1<<k))>0) continue; /// 第k道菜作为当前最后选的菜
                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;
}