不能与直系上司同时参加聚会求最大人数 有根树形DP 最大独立集 poj 3342

http://poj.org/problem?id=3342

Party at Hali-Bula Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5793   Accepted: 2068 Description Dear Contestant, I'm going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I wish I could invite all my co-workers, but imagine how an employee can enjoy a party when he finds his boss among the guests! So, I decide not to invite both an employee and his/her boss. The organizational hierarchy at BCM is such that nobody has more than one boss, and there is one and only one employee with no boss at all (the Big Boss)! Can I ask you to please write a program to determine the maximum number of guests so that no employee is invited when his/her boss is invited too? I've attached the list of employees and the organizational hierarchy of BCM. Best, --Brian Bennett P.S. I would be very grateful if your program can indicate whether the list of people is uniquely determined if I choose to invite the maximum number of guests with that condition. Input The input consists of multiple test cases. Each test case is started with a line containing an integer n (1 ≤ n ≤ 200), the number of BCM employees. The next line contains the name of the Big Boss only. Each of the following n-1 lines contains the name of an employee together with the name of his/her boss. All names are strings of at least one and at most 100 letters and are separated by blanks. The last line of each test case contains a single 0. Output For each test case, write a single line containing a number indicating the maximum number of guests that can be invited according to the required condition, and a word Yes or No, depending on whether the list of guests is unique in that case. Sample Input 6 Jason Jack Jason Joe Jack Jill Jason John Jack Jim Jill 2 Ming Cho Ming 0 Sample Output 4 Yes 1 No Source Tehran 2006

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题目大意：某公司有个聚会，要邀请员工来参加。要求员工和他的直系上司不能同时到这个聚会，问最多能邀请到多少人，有多种邀请方法时输出No

DP过程

注意：对于每个矛盾关系，从老板向员工连一条边

dp[i][0]表示不取i的最大值，可以由两个状态转移而来dp[i][0]+=sigma[ max（dp[j][0],dp[j][1]）],j为儿子，即儿子取或不取都可以

dp[i][1]表示取i的最大值，初始值赋为1，那么儿子节点就不能取了，所以dp[i][1]=sigma(dp[j][0]);

 

 

判断方案是否唯一

 

观察状态转移的过程可知：dp[i][0]是由dp[j][0],dp[j][1]两个状态过来的，所以当dp[j][0]==dp[j][1]时，方案不唯一，即子节点选与不选都可以

但是注意前提是dp[i][0]更优与dp[i][1]，即i这个节点肯定被选择了，否则状态就仅仅由dp[j][1]转移而来，不能判断不唯一。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<map>
using namespace std;
#define max(a,b) a>b?a:b
const int maxn  =  201;
vector<int> edge[maxn];
int dp[210][2];
void dfs(int u,int p)
{
    int i,j;
    dp[u][1]=1;
    dp[u][0]=0;
    for(i=0;i<edge[u].size();i++)
    {
        int v=edge[u][i];
        if(v==p) continue;
        dfs(v,u);
        dp[u][1]+=dp[v][0];
        dp[u][0]+=max(dp[v][0],dp[v][1]);
    }
}
int main(){
     map<string,int> mm;
     int n,i,tot;
     char bos[110],a[110],b[110];
     while(scanf("%d",&n),n)
     {
         tot=0;
         for(i=0;i<=200;i++) edge[i].clear();
         mm.clear();
         scanf("%s",bos);mm[bos]=tot++;
         for(i=0;i<n-1;i++)
         {
             scanf("%s%s",a,b);
             if(mm.find(a)==mm.end()) mm[a]=tot++;
             if(mm.find(b)==mm.end()) mm[b]=tot++;
             edge[mm[b]].push_back(mm[a]);
         }
         dfs(0,0);bool flag=true;
         for(i=0;i<n;i++)
         {
             flag=true;
             if(dp[i][0]>dp[i][1])
             {
                 for(int j=0;j<edge[i].size();j++)
                 {
                     if(dp[edge[i][j]][0]==dp[edge[i][j]][1])
                     {
                         flag=false;
                         break;
                     }
                 }
             }
             if(!flag) break;
         }
         printf("%d",max(dp[0][0],dp[0][1]));
         if(dp[0][0]==dp[0][1]||!flag) printf(" No\n");
         else printf(" Yes\n");
     }
     return 0;
}