Codeforces Round 580 D. Kefa and Dishes （状态压缩dp）

D. 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 ndishes. 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 xexactly 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.

Sample test(s)

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：一共n道菜，吃m道，有k个规则，每个菜有自己的价值，每个规则说明吃完X接着吃Y可以额外获得Z个价值。

问可以获得的最大价值是多少。

思路：1<<18保存所有状态，第二维保存最后吃的哪道菜，然后从吃的菜里选一道，从没吃的菜里选一道，吃完X吃Y，

更新dp，判断一下是否当前吃了m道，更新ans便可，记忆化搜索也可以。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <cmath>using namespace std;const int Max=100005;const int INF=0x3f3f3f3f;long long dp[(1<<18)+10][20];int val[20],add[20][20];int n,m,k;int main(){    int x,y,value,tot,cnt;    long long ans;    while (scanf("%d%d%d",&n,&m,&k)!=EOF)    {                memset(dp, 0, sizeof(dp));        memset(add, 0, sizeof(add));        for (int i=0; i<n; i++)            scanf("%d",&val[i]);        for (int i=0; i<k; i++)        {            scanf("%d%d%d",&x,&y,&value);            add[x-1][y-1]=value;        }                for (int i=0; i<n; i++)            dp[1<<i][i]=val[i];                                tot=1<<n;        ans=0;        for (int i=0; i<tot; i++) //所有状态        {            cnt=0;            for (int j=0; j<n; j++) //已选了第j道            {                                if (i& (1<<j))                {                    cnt++;                    for (int k=0; k<n; k++) //选第k道                    {                        if ((i& (1<<k))==0)                            dp[i | (1<<k)][k]=max(dp[i | (1<<k)][k],dp[i][j]+val[k]+add[j][k]);                    }                }            }            if (cnt==m)            {                for (int j=0; j<n; j++)                {                    if (i& (1<<j))                        ans=max(ans,dp[i][j]);                }            }        }                printf("%lld\n",ans);    }    return 0;}

记忆化：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int Max=100005;const int INF=0x3f3f3f3f;long long dp[(1<<18)+10][20];int val[20],add[20][20];int n,m,k;long long dfs(int x,int y,int z){    if (dp[x][y]!=-1)        return dp[x][y];    dp[x][y]=0;    if (z==m)        return dp[x][y];        for (int i=0; i<n; i++)    {        if (((1 << i) & x) ==0) //当前未选择第i道菜            dp[x][y]=max(dp[x][y],dfs(((1<<i) | x), i, z+1)+val[i]+add[y][i]);    }    return dp[x][y];}int main(){    int x,y,value;    while (scanf("%d%d%d",&n,&m,&k)!=EOF)    {                memset(dp, -1, sizeof(dp));        memset(add, 0, sizeof(add));        for (int i=0; i<n; i++)            scanf("%d",&val[i]);        for (int i=0; i<k; i++)        {            scanf("%d%d%d",&x,&y,&value);            add[x-1][y-1]=value;        }        printf("%lld\n",dfs(0,n+1,0));    }    return 0;}