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hust 1032 Jim

2024-07-04 00:59:33   224人阅读

题目描述

Jim is boss of a big company . There are so many workers in his company that it will take a long time for his command send to all of his workers . One day he want to know how long it will take at least . So he ask you , one of his best programer , to solve the problem . All the workers in Jim‘s conpany will receive command from only one people . So there is a tree (the root is Jim) identify the order Jim‘s command will send by . One send command to another will cost 1 minute and anyone can‘t send command to more than one people at the same time .

输入

There will be multiple cases in the input . For each case there will be just one number N (N<=10000) in the first line , then N-1 fellowed , each line with two names: worker1 worker2 , that is to say worker2 send command to worker1 . The name of the workers will only contain characters and the length is at most 5 .

输出

Just one number to tell how many minutes it will cost at least .

样例输入

5
Tom Jim
Bob Jim
John Tom
Mily Tom

样例输出

3
唉！这个题一直做错，最后是在厚着脸皮问了学长，结果被学长狂虐了，树形dp，膜拜大神他们啊
#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
using namespace std;

vector<int>tree[10001];
map<string,int>people;
typedef map<string,int>::iterator Itr;
Itr itr;
int nowmax;
bool vis[10001];
bool cmp(int x,int y){return x>y;}

int dfs(int u)
{
    int a[10001];
    int sum=tree[u].size();
    if (sum==0) return 0;
    int ans=0,k=0;
    for (int i=0;i<tree[u].size();i++)
    {
        int v=tree[u][i];
        if (!vis[v])
        {
            vis[v]=true;
            a[++k]=dfs(v);
        }
    }
    sort(a+1,a+sum+1,cmp);
    for (int i=1;i<=sum;i++)
    ans=max(ans,a[i]+i);
    return ans;
}

int main()
{
    int n,m;
    string str1,str2;
    m=0;
    while (scanf("%d",&n)!=EOF)
    {
        if (n==1)
        {
            cout<<"0"<<endl;
            continue;
        }
        for (int i=0;i<=n;i++) tree[i].clear();
        m=0;
        people.clear();
        for (int i=1;i<=n-1;i++)
        {
            cin>>str1>>str2;
            if(people.count(str1)==0)
            {
                people[str1]=++m;
                if (people.count(str2)==0)
                {
                    people[str2]=++m;
                    tree[m].push_back(m-1);
                }
                else tree[people[str2]].push_back(m);
            }
            else
            {
                if (people.count(str2)==0)
                {
                    people[str2]=++m;
                    tree[m].push_back(people[str1]);
                }
                else tree[people[str2]].push_back(people[str1]);
            }
        }
        int u=people["Jim"];
        memset(vis,0,sizeof(vis));
        vis[u]=true;
        cout<<dfs(u)<<endl;
    }
    return 0;
}

毕竟是自己写的程序，加油


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