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

Copy
2 2 1
1 1
2 1 1
Output

Copy
3
Input

Copy
4 3 2
1 2 3 4
2 1 5
3 4 2
Output

Copy
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.

题意翻译

kefa进入了一家餐厅,这家餐厅中有n个菜(0<n<=18),kefa对第i个菜的满意度为ai(0<=ai<=10^9),并且对于这n个菜有k个规则,如果kefa在吃完第xi个菜之后吃了第yi个菜(保证xi、yi不相等),那么会额外获得ci(0<=ci<=10^9)的满意度。kefa要吃m道任意的菜(0<m<=n),但是他希望自己吃菜的顺序得到的满意度最大,请你帮帮kefa吧!

输入第一行是三个数:n,m,k

第二行是n个数,第i个数表示kefa对第i道菜的满意度为ai

第三行到第k+2行每行有3个数:xi,yi,ci,表示如果kefa在吃完第xi道菜之后立刻吃了第yi道菜,则会额外获得ci的满意度

Ps:这道题是个状压,因为n,m的范围都很小

n,m很小,那么状压再转移即可;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
} int n, m, k;
ll dp[(1<<18)+6][20];
ll a[maxn];
ll ret[20][20]; int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m); rdint(k);
int cnt = 0;
for (int i = 0; i < n; i++)rdllt(a[i]);
for (int i = 0; i < k; i++) {
ll x, y, z; rdllt(x); rdllt(y); rdllt(z);
ret[x - 1][y - 1] = z;
}
for (int i = 0; i < n; i++) {
dp[1 << (i)][i] = a[i];
}
ll maxx = -1;
for (int i = 0; i < (1 << n); i++) {// 枚举状态
cnt = 0;
for (int j = 0; j < n; j++) {
if (i&(1 << j)) {// 看该位是否为1
cnt++;
for (int k = 0; k < n; k++) {// 选择吃下一个
if (!(i&(1 << k))) {
dp[i | (1 << k)][k] = max(dp[i|(1 << k)][k], dp[i][j] + a[k] + ret[j][k]);
// dp 转移
}
}
}
}
if (cnt == m) {// 满足 m
for (int j = 0; j < n; j++)
if (i&(1 << j))maxx = max(maxx, dp[i][j]);
}
}
cout << (ll)maxx << endl;
return 0;
}

CF580D Kefa and Dishes 状压dp的更多相关文章

  1. 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 W ...

  2. Codeforces ----- Kefa and Dishes [状压dp]

    题目传送门:580D 题目大意:给你n道菜以及每道菜一个权值,k个条件,即第y道菜在第x道后马上吃有z的附加值,求从中取m道菜的最大权值 看到这道题,我们会想到去枚举,但是很显然这是会超时的,再一看数 ...

  3. codeforces 580D Kefa and Dishes(状压dp)

    题意:给定n个菜,每个菜都有一个价值,给定k个规则,每个规则描述吃菜的顺序:i j w,按照先吃i接着吃j,可以多增加w的价值.问如果吃m个菜,最大价值是多大.其中n<=18 思路:一看n这么小 ...

  4. Codeforces Round #321 (Div. 2) Kefa and Dishes 状压+spfa

    原题链接:http://codeforces.com/contest/580/problem/D 题意: 给你一些一个有向图,求不超过m步的情况下,能获得的最大权值和是多少,点不能重复走. 题解: 令 ...

  5. D. Kefa and Dishes(状压)

    永久打开的传送门 \(这次总算没有写砸........\) \(设f[i][j]为上一次吃的i物品状态为j的最大收益\) \(那么我们就暴力枚举所有状态i,然后在当前状态找出一个没吃的食物j,再去找一 ...

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

    http://codeforces.com/contest/580/problem/D 题意: 有个人去餐厅吃饭,现在有n个菜,但是他只需要m个菜,每个菜只吃一份,每份菜都有一个欢乐值.除此之外,还有 ...

  7. 【62测试】【状压dp】【dfs序】【线段树】

    第一题: 给出一个长度不超过100只包含'B'和'R'的字符串,将其无限重复下去. 比如,BBRB则会形成 BBRBBBRBBBRB 现在给出一个区间[l,r]询问该区间内有多少个字符'B'(区间下标 ...

  8. 状压dp入门

    状压dp的含义 在我们解决动态规划题目的时候,dp数组最重要的一维就是保存状态信息,但是有些题目它的具有dp的特性,并且状态较多,如果直接保存的可能需要三维甚至多维数组,这样在题目允许的内存下势必是开 ...

  9. [CF580D]Kefa and Dishes

    题意翻译 kefa进入了一家餐厅,这家餐厅中有n个菜(0<n<=18),kefa对第i个菜的满意度为ai(0<=ai<=10^9),并且对于这n个菜有k个规则,如果kefa在吃 ...

随机推荐

  1. CreateMutex实现只能打开一个客户端

    #include "stdafx.h" #include <Windows.h> #include <iostream> using namespace s ...

  2. paramiko分开执行多条命令 不像之前一样使用\n

    #!/usr/bin/env python#-*- encoding -*- import paramiko transport = paramiko.Transport(('192.168.11.1 ...

  3. 问题:sqlserver 跨服务器连接;结果:Sql Server 跨服务器连接

    Sql Server 跨服务器连接 用openrowset连接远程SQL或插入数据 --如果只是临时访问,可以直接用openrowset --查询示例 select * from openrowset ...

  4. 问题:只能在执行 Render() 的过程中调用 RegisterForEventValidation;结果:只能在执行 Render() 的过程中调用 RegisterForEventValidation

    只能在执行 Render() 的过程中调用 RegisterForEventValidation 当在导出Execl或Word的时候,会发生只能在执行 Render() 的过程中调用 Register ...

  5. javascript删除option选项的多种方法总结

    转自:https://blog.csdn.net/xiaoxuanyunmeng/article/details/16886505 1. JavaScript 代码如下: var oSel=docum ...

  6. C#如何生成JSON字符串?(序列化对象)

    第一章:C#如何拿到从http上返回JSON数据? 第二章:C#如何解析JSON数据?(反序列化对象) 第三章:C#如何生成JSON字符串?(序列化对象) 第四章:C#如何生成JSON字符串提交给接口 ...

  7. Solr根据参考点的坐标来返回范围内的小区和距离

    @Test public void query() throws Exception{ SystemDefaultHttpClient httpClient = new SystemDefaultHt ...

  8. Java开源中文分词类库

      IKAnalyzer  IKAnalyzer是一个开源的,基于java语言开发的轻量级的中文分词工具包.从2006年12月推出1.0版开始,IKAnalyzer已经推出了3个大版本.最初,它是以开 ...

  9. HDU 6396(2018多校第七场1011) Swordsman

    场上场下各种TLE到怀疑人生...经过大佬指点之后才知道要用fread才能过,一般的快读不行... 题意:一个剑客打小怪兽,有n头小怪兽,剑客和小怪兽有m个属性.只有剑客的m个属性都大于等于某个小怪兽 ...

  10. ???Struts2框架03 session的使用、登录逻辑【session工作原理】

    1 登录逻辑 1.1 获取登录数据(例如:用户名.密码) 1.2 在控制层调用业务层来验证数据信息 1.3 登录成功:保存用户信息(服务器用session.浏览器用cookie),跳转到主页面 1.4 ...