HDU2412 樹狀DP

Party at Hali-Bula

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 138    Accepted Submission(s): 51

Problem 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

 

給定一棵關係樹 , 從中選擇一些點 , 使這些點均不存在親子關係 , 最多能取多少個點 , 並且判斷取法是否唯一 .

經典的樹狀 DP.

設 dp[i][0] 爲在以 i 爲根的子樹中 , 不選擇點 i 最多能夠選的數目 ,dp[i][1] 爲選擇 i 點的最多數目 .

狀態轉移方程 :

當 i 爲葉子節點時 :

dp[i][0]=0;

dp[i][1]=1;

當 i 爲非葉子節點時 :

dp[i][0]=sum(max(dp[j][0],dp[j][1]))  (j 爲 i 的兒子 )

dp[i][1]=sum(dp[j][0])  (j 爲 i 的兒子 )

 

解的唯一性的判斷 :

設 u[i][x] 爲 0 時表示 dp[i][x] 的解唯一 , 爲 1 則表示不唯一 .

當 x 爲 0 時 , 若存在 j 是 i 的兒子 , 使得 dp[j][0]>dp[j][1] 且 u[j][0]=1, 或 dp[j][0]<dp[j][1] 且 u[j][1]=1 或 dp[j][0]=dp[j][1], 則 u[i][0]=1;

當 x 爲 1 時 , 若存在 j 是 i 的兒子 , 使得 u[j][0]=1, 則 u[i][0]=1;

 

代碼如下 :

#include<stdio.h> #include<string.h> #define MAXN 205 struct Node { char name[105]; Node *next; }node[MAXN]; struct List { int node; List *next; }head[MAXN]; int dp[MAXN][2],u[MAXN][2]; void DP(int node) { List *p; int sumdp=0; if (head[node].next==NULL) { dp[node][0]=0; dp[node][1]=1; u[node][0]=0; u[node][1]=0; } else { p=head[node].next; while (p!=NULL) { DP(p->node); if (dp[p->node][0]>dp[p->node][1]) { sumdp+=dp[p->node][0]; if (u[p->node][0]==1) u[node][0]=1; } else if (dp[p->node][0]<dp[p->node][1]) { sumdp+=dp[p->node][1]; if (u[p->node][1]==1) u[node][0]=1; } else { sumdp+=dp[p->node][0]; u[node][0]=1; } dp[node][1]+=dp[p->node][0]; if (u[p->node][0]==1) u[node][1]=1; p=p->next; } dp[node][0]=sumdp; dp[node][1]+=1; } } int main() { List *s; int i,j,n; char s1[MAXN][105],s2[MAXN][105]; while (scanf("%d",&n)&&n) { memset(dp,0,sizeof(dp)); memset(u,0,sizeof(u)); memset(s1,0,sizeof(s1)); memset(s2,0,sizeof(s2)); for (i=0;i<=n;++i) head[i].next=NULL; getchar(); scanf("%s",node[0].name); getchar(); for (i=1;i<n;++i) { scanf("%s %s",s1[i],s2[i]); getchar(); strcpy(node[i].name,s1[i]); } for (i=1;i<n;++i) { for (j=0;j<n;++j) { if (strcmp(s2[i],node[j].name)==0) { s=new List; s->node=i; s->next=head[j].next; head[j].next=s; break; } } } DP(0); if (dp[0][0]>dp[0][1]) { printf("%d ",dp[0][0]); if (u[0][0]) printf("No/n"); else printf("Yes/n"); } else if (dp[0][0]<dp[0][1]) { printf("%d ",dp[0][1]); if (u[0][1]) printf("No/n"); else printf("Yes/n"); } else printf("%d No/n",dp[0][0]); } return 0; }