ACM&&树形DP

本文介绍了一种解决特定战略游戏问题的方法,该游戏的目标是在一棵树形结构的城市中部署最少数量的士兵来监视所有道路。文章详细阐述了使用动态规划算法进行求解的过程,并提供了完整的代码实现。
Strategic game
Time Limit: 2000MS
Memory Limit: 10000K
Total Submissions: 3555
Accepted: 1556

Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

For example for the tree:

the solution is one soldier ( at the node 1).

Input

The input contains several data sets in text format. Each data set represents a tree with the following description:

  • the number of nodes
  • the description of each node in the following format
    node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifiernumber_of_roads

    or
    node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500);the number_of_roads in each line of input will no more than 10. Every edge appears only once in the input data.

Output

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following:

Sample Input

4
0:(1) 1
1:(2) 2 3
2:(0)
3:(0)
5
3:(3) 1 4 2
1:(1) 0
2:(0)
0:(0)
4:(0)

Sample Output

1
2

Source

Southeastern Europe 2000


/*
ans[i][0]表示在不选择节点i的情况下,以i为根节点的子树,最少需要选择的点数;

ans[i][1]表示在选择节点i的情况下,以i为根节点的子树,最少需要选择的点数;

当i是叶子时,ans[i][0]=0,ans[i][1]=1;

else

ans[i][0]=sigma(ans[j][1])(j为i的子节点)

ans[i][1]=1+sigmamin(ans[j][0],ans[j][1]);
 */

#include<iostream>
#include<cstdio>
using namespace std;
int min(int a,int b)
{
  if(a>b) return b;
  else return a;
}
int dp[1505][2];
int child[1505];
int pos[1505];
struct EDGE
{
  int from,to;
}edge[1505];
void FindDp(int st)//递归时是把st压入栈中的,不是to
{
  int to;
  dp[st][0] = 0;                                                                
  dp[st][1] = 1;                                      
  if(child[st] == 0) return ;//当递归到叶子节点就返回
  int i = pos[st];
  while(edge[i].from == st)
    {
      to = edge[i++].to;
      FindDp(to);//从儿子节点继续递归搜
      dp[st][0] += dp[to][1];
      dp[st][1] += min(dp[to][0],dp[to][1]);
    }
}
int main()
{
  int n;
  int parent,sum;
  int num;
  int t,count,level;
  while(scanf("%d",&n) != EOF)
    {
      num = level = 0;
      count = 1;
      for(int j =1;j<=n;j++)
    {
      scanf("%d:(%d)",&parent,&sum);
      num += parent;
      pos[parent] = count;
      child[parent] = sum;
      for(int i =1;i <= sum;i++)
        {
          scanf("%d",&t);
          level += t;
          edge[count].from = parent;
          edge[count++].to = t;
        }
    }
      int find = num - level;//记录根节点
      FindDp(find);//从根节点开始递归
      printf("%d\n",min(dp[find][0],dp[find][1]));
    }
}



内容概要:本文系统介绍了算术优化算法(AOA)的基本原理、核心思想及Python实现方法,并通过图像分割的实际案例展示了其应用价值。AOA是一种基于种群的元启发式算法,其核心思想来源于四则运算,利用乘除运算进行全局勘探,加减运算进行局部开发,通过数学优化器加速函数(MOA)和数学优化概率(MOP)动态控制搜索过程,在全局探索与局部开发之间实现平衡。文章详细解析了算法的初始化、勘探与开发阶段的更新策略,并提供了完整的Python代码实现,结合Rastrigin函数进行测试验证。进一步地,以Flask框架搭建前后端分离系统,将AOA应用于图像分割任务,展示了其在实际工程中的可行性与高效性。最后,通过收敛速度、寻优精度等指标评估算法性能,并提出自适应参数调整、模型优化和并行计算等改进策略。; 适合人群:具备一定Python编程基础和优化算法基础知识的高校学生、科研人员及工程技术人员,尤其适合从事人工智能、图像处理、智能优化等领域的从业者;; 使用场景及目标:①理解元启发式算法的设计思想与实现机制;②掌握AOA在函数优化、图像分割等实际问题中的建模与求解方法;③学习如何将优化算法集成到Web系统中实现工程化应用;④为算法性能评估与改进提供实践参考; 阅读建议:建议读者结合代码逐行调试,深入理解算法流程中MOA与MOP的作用机制,尝试在不同测试函数上运行算法以观察性能差异,并可进一步扩展图像分割模块,引入更复杂的预处理或后处理技术以提升分割效果。
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