Anniversary party
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9976 Accepted Submission(s): 4234
Problem Description
There
is going to be a party to celebrate the 80-th Anniversary of the Ural
State University. The University has a hierarchical structure of
employees. It means that the supervisor relation forms a tree rooted at
the rector V. E. Tretyakov. In order to make the party funny for every
one, the rector does not want both an employee and his or her immediate
supervisor to be present. The personnel office has evaluated
conviviality of each employee, so everyone has some number (rating)
attached to him or her. Your task is to make a list of guests with the
maximal possible sum of guests' conviviality ratings.
Input
Employees
are numbered from 1 to N. A first line of input contains a number N. 1
<= N <= 6 000. Each of the subsequent N lines contains the
conviviality rating of the corresponding employee. Conviviality rating
is an integer number in a range from -128 to 127. After that go T lines
that describe a supervisor relation tree. Each line of the tree
specification has the form:
L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line
0 0
L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line
0 0
Output
Output should contain the maximal sum of guests' ratings.
Sample Input
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0
Sample Output
5
思路：树形dp;
dp[i][j] 表示第j个节点状态为i时i ->>[0,1]该节点选或不选，然后从根节点dfs dp就可以了。复杂度O(n);
1 #include<stdio.h> 2 #include<math.h> 3 #include<queue> 4 #include<algorithm> 5 #include<string.h> 6 #include<iostream> 7 #include<stack> 8 #include<vector> 9 using namespace std; 10 typedef long long LL; 11 int val[7000]; 12 vector<int>vec[7000]; 13 int dp[2][7000]; 14 int cnt[7000]; 15 void dfs(int n); 16 int main(void) 17 { 18 int n; 19 while(scanf("%d",&n)!=EOF) 20 { 21 int i,j; 22 memset(cnt,0,sizeof(cnt)); 23 for(i = 0; i < 7000; i++) 24 vec[i].clear(); 25 memset(dp,0,sizeof(dp)); 26 for(i = 1; i <= n; i++) 27 scanf("%d",&val[i]); 28 int ch,fa; 29 while(scanf("%d %d",&ch,&fa),ch!=0&&fa!=0) 30 { 31 cnt[ch] = 1; 32 vec[fa].push_back(ch); 33 } 34 int root; 35 for(i = 1; i <= n; i++) 36 { 37 if(!cnt[i]) 38 { 39 root = i; 40 break; 41 } 42 } 43 dfs(root); 44 if(dp[1][root]<0&&dp[0][root]<0) 45 { 46 printf("0 "); 47 } 48 else 49 { 50 int ask = max(dp[1][root],dp[0][root]); 51 printf("%d ",ask); 52 } 53 } 54 return 0; 55 } 56 void dfs(int n) 57 { 58 int sum1 = 0; 59 int sum2 = 0; 60 for(int i = 0; i < vec[n].size(); i++) 61 { 62 int id = vec[n][i]; 63 dfs(id); 64 sum1 += dp[0][id]; 65 sum2 += max(max(dp[1][id],dp[0][id]),0); 66 } 67 dp[0][n] = max(dp[0][n],sum2); 68 dp[0][n] = max(dp[0][n],sum1); 69 dp[1][n] = max(dp[0][n],val[n]); 70 dp[1][n] = max(dp[0][n],sum1+val[n]); 71 }