poj-2342-简单树形dp
Anniversary party
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10388 | Accepted: 5943 |
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 N – 1 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
Source
给出一颗有根树，每个点有权值，挑选一些点使权值和达到最大，如果选择了u点，则u的儿子便不能再选了，但是儿子下面的节点还可以选。
直接树形dp，f[i][0]表示不选i达到的最大值，f[i][1]表示选了i点能达到的最大值，贪心的选择正数的子树。
1 #include<iostream> 2 #include<bitset> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 #define inf 0x3f3f3f3f 7 int fa[6060]; 8 int son[6060]; 9 int next[6060]; 10 int f[6060][2]; 11 int c[6060]; 12 int N,root; 13 void dfs(int u,int father){ 14 f[u][1]=c[u]; 15 f[u][0]=0; 16 int s=0; 17 for(int i=son[u];i;i=next[i]){ 18 if(i==father) continue; 19 dfs(i,u); 20 if(f[i][0]>0) f[u][1]+=f[i][0]; 21 f[u][0]+=max(0,max(f[i][0],f[i][1])); 22 } 23 } 24 int main() 25 { 26 int i,j,k; 27 int u,v; 28 while(cin>>N){ 29 if(N==0){ 30 cin>>N; 31 break; 32 } 33 memset(fa,0,sizeof(fa)); 34 memset(son,0,sizeof(son)); 35 memset(next,0,sizeof(next)); 36 for(i=1;i<=N;++i) scanf("%d",c+i); 37 for(i=1;i<N;++i){ 38 scanf("%d%d",&u,&v); 39 fa[u]=v; 40 next[u]=son[v]; 41 son[v]=u; 42 } 43 for(i=1;i<=N;++i) 44 if(!fa[i]) {root=i;break;} 45 dfs(root,0); 46 cout<<max(f[root][0],f[root][1])<<endl; 47 } 48 return 0; 49 }
