CF580D Kefa and Dishes 状压dp
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!
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.
In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.
2 2 1
1 1
2 1 1
3
4 3 2
1 2 3 4
2 1 5
3 4 2
12
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 【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 ...
- 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 ...
- [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
- 2015.3.2 VC++6制作非MFC dll以及VS2005、VS2010调用
1.在VC6中新建工程,选择Win32 Dynamic-Link Libary,输入dll名称如 DLL2015 2.在类型选择中,选择第2项 A Simple Dll project OK 3.随后 ...
- PHP中交换两个变量的值
首先,采用php的list数据结构.上代码,然后再解析 function swap(&$a, &$b) { list ( $a, $b ) = array ($b, $a ); } l ...
- mongodb用mongoose得到的对象不能增加属性解决
一,先定义了一个goods(商品)的models var mongoose = require('mongoose'); var Schema = mongoose.Schema; var produ ...
- spring的配置文件在web.xml中加载的方式
web.xml加载spring配置文件的方式主要依据该配置文件的名称和存放的位置不同来区别,目前主要有两种方式. 1.如果spring配置文件的名称为applicationContext.xml,并且 ...
- form(去掉关闭按钮,禁止调整大小)
禁止Form窗口调整大小方法:FormBorderStyle 设为FixedSingle: 不能使用最大化窗口: MaximuzeBox 设为False: 不能使用最小化窗口: MinimizeB ...
- docker 笔记 (7) 限制容器
内存 -m 或 --memory:设置内存的使用限额,例如 100M, 2G.--memory-swap:设置 内存+swap 的使用限额.--vm 1:启动 1 个内存工作线程.--vm-bytes ...
- LAMP 3.0 mysql配置讲解
mysql 安装好后,我们是从安装包的 support-files 里面复制过来一个模板配置文件,默认 mysql 配置文件是在/etc/my.cnf 下,其实这个路径或者文件名字我们是可以修改的,在 ...
- SQL查询语句 [1]
一.使用字符串作为条件查询 在 Home/controller/UserController.class.php 下插入 <?php namespace Home\Controller; use ...
- ROS Learning-002 beginner_Tutorials 如何添加ROS环境变量 和 如何更新ROS源代码
ROS Indigo beginner_Tutorials 之 添加环境变量 和 更新ROS源代码的命令 我使用的虚拟机软件:VMware Workstation 11 使用的Ubuntu系统:Ubu ...
- 后台执行UNIX/Linux命令和脚本的五种方法
hiveserver 后台启动 nohup "${HIVE_HOME}"/bin/hive --service hiveserver2 & 1. 使用&符号在后台执 ...