codeforces 580D. Kefa and Dishes

本文介绍了一个使用状态压缩动态规划解决的点餐问题,目标是在遵循特定的菜品搭配规则下,从有限的菜品中选择若干份以获得最大的满意度。文章详细解释了解决方案的实现思路和代码细节。

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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 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, nm and k (1 ≤ m ≤ n ≤ 180 ≤ 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 xiyi and ci (1 ≤ xi, yi ≤ n0 ≤ 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 indexesi 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 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.


这题可以用状压dp做,用dp[state][j]表示取了state里的1的这些菜,最后取的菜是j最多能得到的价值。状态转移方程是dp[state1][j]=max(dp[state1][j],dp[state][i]+a[i][j]+value[j] );

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef __int64 ll;
#define inf 99999999
#define pi acos(-1.0)
ll dp[280000][20],value[20];
ll a[20][20];

int panduan(int state,int m)
{
    int tot=0;
    while(state){
        if(state&1)tot++;
        state>>=1;
    }
    if(tot==m)return 1;
    return 0;
}
int main()
{
    int i,j,c,d,k,n,m,state,state1;
    ll e;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(a,0,sizeof(a));
        for(i=1;i<=n;i++){
            scanf("%I64d",&value[i]);
        }
        for(i=1;i<=k;i++){
            scanf("%d%d%I64d",&c,&d,&e);
            a[c][d]=e;
        }
        ll maxnum=0;
        if(m==1){
            for(i=1;i<=n;i++){
                maxnum=max(maxnum,value[i]);
            }
            printf("%I64d\n",maxnum);
            continue;
        }

        for(state=1;state<=(1<<n)-1;state++){
            for(i=1;i<=n;i++){
                if(state&(1<<(i-1))){
                    if(state==(1<<(i-1) )){
                        dp[state][i]=value[i];
                    }
                    for(j=1;j<=n;j++){
                        if( (state&(1<<(j-1) ) )==0){
                            state1=( state|(1<<(j-1)) );
                            dp[state1][j]=max(dp[state1][j],dp[state][i]+a[i][j]+value[j] );
                        }
                    }

                }
            }
        }

        maxnum=0;
        for(state=1;state<=(1<<n)-1;state++){
            if(panduan(state,m)){
                for(i=1;i<=n;i++){
                    maxnum=max(maxnum,dp[state][i]);
                }
            }
        }
        printf("%I64d\n",maxnum);
    }
    return 0;
}




### Codeforces Div.2 比赛难度介绍 Codeforces Div.2 比赛主要面向的是具有基础编程技能到中级水平的选手。这类比赛通常吸引了大量来自全球不同背景的参赛者,包括大学生、高中生以及一些专业人士。 #### 参加资格 为了参加 Div.2 比赛,选手的评级应不超过 2099 分[^1]。这意味着该级别的竞赛适合那些已经掌握了一定算法知识并能熟练运用至少一种编程语言的人群参与挑战。 #### 题目设置 每场 Div.2 比赛一般会提供五至七道题目,在某些特殊情况下可能会更多或更少。这些题目按照预计解决难度递增排列: - **简单题(A, B 类型)**: 主要测试基本的数据结构操作和常见算法的应用能力;例如数组处理、字符串匹配等。 - **中等偏难题(C, D 类型)**: 开始涉及较为复杂的逻辑推理能力和特定领域内的高级技巧;比如图论中的最短路径计算或是动态规划入门应用实例。 - **高难度题(E及以上类型)**: 对于这些问题,则更加侧重考察深入理解复杂概念的能力,并能够灵活组合多种方法来解决问题;这往往需要较强的创造力与丰富的实践经验支持。 对于新手来说,建议先专注于理解和练习前几类较容易的问题,随着经验积累和技术提升再逐步尝试更高层次的任务。 ```cpp // 示例代码展示如何判断一个数是否为偶数 #include <iostream> using namespace std; bool is_even(int num){ return num % 2 == 0; } int main(){ int number = 4; // 测试数据 if(is_even(number)){ cout << "The given number is even."; }else{ cout << "The given number is odd."; } } ```
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