POJ 3342 Party at Hali-Bula (树形dp 树的最大独立集判多解好题)-白红宇

POJ 3342 Party at Hali-Bula (树形dp 树的最大独立集判多解好题)

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发布时间：2019-06-29

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Party at Hali-Bula

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5660		Accepted: 2022

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

6JasonJack JasonJoe JackJill JasonJohn JackJim Jill2MingCho Ming0

Sample Output

4 Yes1 No

Source

题目链接：

题目大意：一棵树，父亲和儿子不能同一时候选入同一个集合，如今求能选集合中元素个数最多的那个集合大小。并推断解是否唯一

题目分析：求树的最大独立集的问题。好题，也用了两次dp的思想，有点类似这题的思想

dp[i][0]表示不选第i个结点，集合大小的最大值

dp[i][1]表示选第i个结点，集合大小的最大值

对于此dp显然

dp[i][1] = dp[son][0] 选父亲则不能选儿子

dp[i][0] = max(dp[son][0], dp[son][1]) 不选父亲的话则值等于选儿子或者不选儿子里的较大值

s[i][1] == true 表示选第i个结点时有唯一解，false表示解不唯一

s[i][0] == true 表示不选第i个结点时有唯一解。false表示解不唯一

開始时设s[i][1]，s[i][0]都为true，对于此dp。我们主要考虑父亲的解变成不唯一的情况

与第一个dp状态相应，分成选父亲和不选父亲两种情况

if(!s[son][0]) s[i][1] = false，意思是假设不选儿子时有多个解，则此时能够选父亲，选父亲也肯定有多个解

if(dp[son][0] == dp[son][1]) s[i][0] = false。假设选不选儿子的答案同样。显然不选父亲时有多个解，由于选不选儿子都能够

最后自叶子向根回溯求解推断就可以。这题由于map没清零，wa了大半天。

。。

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             using namespace std;int const MAX = 300;int n;int dp[MAX][2];bool s[MAX][2];bool vis[MAX];vector 
             
               vt[MAX];void DFS(int fa){ dp[fa][1] = 1; vis[fa] = true; s[fa][0] = true; s[fa][1] = true; int sz = vt[fa].size(); for(int i = 0; i < sz; i++) { int son = vt[fa][i]; if(!vis[son]) { DFS(son); dp[fa][1] += dp[son][0]; dp[fa][0] += max(dp[son][0], dp[son][1]); if(dp[son][0] == dp[son][1]) s[fa][0] = false; if(!s[son][0]) s[fa][1] = false; } } return;}int main(){ while(scanf("%d", &n) != EOF && n) { map 
              
                mp; for(int i = 0; i < MAX; i++) vt[i].clear(); memset(vis, false, sizeof(vis)); memset(dp, 0, sizeof(dp)); int cnt = 0; string boss, fir, sec; cin >> boss; mp[boss] = cnt ++; for(int i = 0; i < n - 1; i++) { cin >> sec >> fir; if(!mp.count(fir)) mp[fir] = cnt ++; if(!mp.count(sec)) mp[sec] = cnt ++; vt[mp[fir]].push_back(mp[sec]); } DFS(0); if(dp[0][1] > dp[0][0] && s[0][1]) printf("%d Yes\n", dp[0][1]); else if(dp[0][1] < dp[0][0] && s[0][0]) printf("%d Yes\n", dp[0][0]); else printf("%d No\n", max(dp[0][0] , dp[0][1])); }}

转载于:https://www.cnblogs.com/gavanwanggw/p/6851549.html

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