Codeforces Round #321 (Div. 2) D. Kefa and Dishes 状压dp

题目链接：

题目

D. Kefa and Dishes

time limit per test：2 seconds

memory limit per test：256 megabytes

问题描述

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!

输入

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.

输出

In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.

样例

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

Nodte

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，每到菜有一个满意度，并且有k个关系，如果在吃了第x道菜之后马上吃第y道，就会增加额外的满意度。现在问你要怎么吃才能使满意度最高。

题解

如果是吃n道菜，就是赤裸裸的状压dp了，不过其实只吃m道菜相当于输出中间过程啦，做完n道菜的之后找dp[i][j]中i的1的个数为m的，更新答案。

（一开始竟然想先从n道中选出m道，然后做状压。。orz）

dp[i][j]表示已经吃了状态i的菜，且最后一次吃的是j的能得到的最大满意度。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define lson (o<<1)
#define rson ((o<<1)|1)
#define M l+(r-l)/2
using namespace std;
const int maxn=20;
typedef __int64 LL;
int n,m,k;
LL dp[1<<maxn][maxn];
int arr[maxn],mat[maxn][maxn];
int main() {
	memset(mat,0,sizeof(mat));
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0; i<n; i++) {
		scanf("%d",&arr[i]);
	}
	while(k--) {
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w),u--,v--;
		mat[v][u]=w;
	}
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;i++){
		dp[1<<i][i]=arr[i];
	}
	LL ans1=-1;
	for(int i=1; i<(1<<n); i++) {
		for(int j=0; j<n; j++) {
			if(i&(1<<j)) {
				for(int k=0; k<n; k++) {
					if(k==j) continue;
					if(i&(1<<k)) {
						dp[i][j]=max(dp[i][j],dp[i^(1<<j)][k]+mat[j][k]+arr[j]);
					}
				}
			}
		}
	}
	LL ans=0;
	for(int i=0; i<(1<<n); i++) {
		int cnt=0;
		for(int k=0; k<n; k++) {
			if(i&(1<<k)) cnt++;
		}
		if(cnt==m) {
			for(int j=0; j<n; j++) {
				if(i&(1<<j)){
					ans=max(ans,dp[i][j]);
				}
			}
		}
	}
	printf("%I64d\n",ans);
	return 0;
}