zoukankan      html  css  js  c++  java
  • POJ1463(树形DP)

    Strategic game
    Time Limit: 2000MS   Memory Limit: 10000K
    Total Submissions: 7487   Accepted: 3482

    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

    没有上司的舞会问题

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    const int MAXN=1505;
    vector<int> tree[MAXN];
    int n;
    int dp[MAXN][2];
    void dfs(int u)
    {
        for(int i=0;i<tree[u].size();i++)
        {
            int v=tree[u][i];
            dfs(v);
            dp[u][0]+=dp[v][1];//不选自己,则必须选子节点
            dp[u][1]+=min(dp[v][1],dp[v][0]);//选上自己,子节点选不选挑个最小的
        }
        dp[u][1]+=1;
    }
    int deg[MAXN];
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            memset(dp,0,sizeof(dp));
            memset(deg,0,sizeof(deg));
            for(int i=0;i<n;i++)
                tree[i].clear();
            for(int i=0;i<n;i++)
            {
                int f,m,v;
                scanf("%d%*c%*c%d%*c",&f,&m);
                while(m--)
                {    
                    scanf("%d",&v);
                    tree[f].push_back(v);
                    deg[v]++;
                }
            }
            for(int i=0;i<n;i++)
            {
                if(deg[i]==0)
                {
                    dfs(i);
                    printf("%d
    ",min(dp[i][0],dp[i][1]));
                    break;    
                }
            }
        }
    }


  • 相关阅读:
    Netbeans 注释模板配置
    你可能不知道的5 个强大的HTML5 API 函数
    科幻大片中那些牛X代码真相
    怎么才能成为一名PHP专家?
    网页设计中常见的错误列举
    五个必须警惕的数据库设计错误
    五种情况会导致Session 丢失
    四种策略防止用户将表单重复提交
    jQuery的deferred对象使用笔记
    [Web前端]由cookies安全说开去
  • 原文地址:https://www.cnblogs.com/program-ccc/p/5222464.html
Copyright © 2011-2022 走看看