【HDU-5000】Clone(dp)

在食用了来自切尔诺贝利的食物后,DRD获得了克隆自己的超能力。然而,这种力量并不完美,克隆体的能力各不其一。为了确保尽可能多的克隆体能够生存下来,ATM试图找出在考虑到每个克隆体N种能力的情况下,最多能有多少个克隆体共存。通过使用动态规划策略,将问题转化为寻找属性总和达到一半的组合,以确定最大生存克隆体的数量。

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Problem Description
After eating food from Chernobyl, DRD got a super power: he could clone himself right now! He used this power for several times. He found out that this power was not as perfect as he wanted. For example, some of the cloned objects were tall, while some were short; some of them were fat, and some were thin.

More evidence showed that for two clones A and B, if A was no worse than B in all fields, then B could not survive. More specifically, DRD used a vector v to represent each of his clones. The vector v has n dimensions, representing a clone having N abilities. For the i-th dimension, v[i] is an integer between 0 and T[i], where 0 is the worst and T[i] is the best. For two clones A and B, whose corresponding vectors were p and q, if for 1 <= i <= N, p[i] >= q[i], then B could not survive.

Now, as DRD’s friend, ATM wants to know how many clones can survive at most.

Input
The first line contains an integer T, denoting the number of the test cases.

For each test case: The first line contains 1 integer N, 1 <= N <= 2000. The second line contains N integers indicating T[1], T[2], …, T[N]. It guarantees that the sum of T[i] in each test case is no more than 2000 and 1 <= T[i].

Output
For each test case, output an integer representing the answer MOD 10^9 + 7.

Sample Input
2
1
5
2
8 6

Sample Output
1
7
有点类似于分组dp,我们发现如果我们使得每个人的属性值的和达到所有总属性和的一半,那么我们可以保证每个人都可以存活而且数量最多,然后就是去dp了
dp[i][j]dp[i][j]为考虑了前i个属性,和达到j时的种类数。
代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
ll dp[2005][2005];
int T[2005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        int n;
        scanf("%d",&n);
        int sum=0;
        for(int i=1;i<=n;i++)scanf("%d",T+i),sum+=T[i];
        sum/=2;
        dp[0][0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=sum;j++)
                for(int k=0;k<=T[i];k++)
                {
                    if(k>j)break;
                    dp[i][j]=(dp[i][j]+dp[i-1][j-k])%mod;
                }
        }
        printf("%lld\n",dp[n][sum]);
    }
    return 0;
}
### 关于HDU - 6609 的题目解析 由于当前未提供具体关于 HDU - 6609 题目的详细描述,以下是基于一般算法竞赛题型可能涉及的内容进行推测和解答。 #### 可能的题目背景 假设该题目属于动态规划类问题(类似于多重背包问题),其核心在于优化资源分配或路径选择。此类问题通常会给出一组物品及其属性(如重量、价值等)以及约束条件(如容量限制)。目标是最优地选取某些物品使得满足特定的目标函数[^2]。 #### 动态转移方程设计 如果此题确实是一个变种的背包问题,则可以采用如下状态定义方法: 设 `dp[i][j]` 表示前 i 种物品,在某种条件下达到 j 值时的最大收益或者最小代价。对于每一种新加入考虑范围内的物体 k ,更新规则可能是这样的形式: ```python for i in range(n): for s in range(V, w[k]-1, -1): dp[s] = max(dp[s], dp[s-w[k]] + v[k]) ``` 这里需要注意边界情况处理以及初始化设置合理值来保证计算准确性。 另外还有一种可能性就是它涉及到组合数学方面知识或者是图论最短路等相关知识点。如果是后者的话那么就需要构建相应的邻接表表示图形结构并通过Dijkstra/Bellman-Ford/Floyd-Warshall等经典算法求解两点间距离等问题了[^4]。 最后按照输出格式要求打印结果字符串"Case #X: Y"[^3]。 #### 示例代码片段 下面展示了一个简单的伪代码框架用于解决上述提到类型的DP问题: ```python def solve(): t=int(input()) res=[] cas=1 while(t>0): n,k=list(map(int,input().split())) # Initialize your data structures here ans=find_min_unhappiness() # Implement function find_min_unhappiness() res.append(f'Case #{cas}: {round(ans)}') cas+=1 t-=1 print("\n".join(res)) solve() ```
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