hdu 1520，poj2342

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12664    Accepted Submission(s): 5106

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

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

当前点有两个状态，一个是去dp[u][1]一个不去dp[u][0]。

所以dp[u][1]=dp[uson1][0]+dp[uson2][0]+...dp[usonn][0] uson是u的孩子。

dp[u][0]=max(dp[uson1][0],dp[uson1][1])+max(dp[uson2][0],dp[uson2][1])+...max(dp[usonn][0],dp[usonn][1])，同样的概念。

难的是杭电的读入会坑人。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define clr(x) memset(x,0,sizeof(x))
 6 #define clr_1(x) memset(x,-1,sizeof(x))
 7 using namespace std;
 8 const int N=1e4+10;
 9 struct edg
10 {
11     int next,to;
12 }edge[N*2];
13 int head[N],ecnt;
14 int dp[N][2],value[N],fa[N];
15 void addedge(int u,int v)
16 {
17     edge[++ecnt]=(edg){head[u],v};
18     head[u]=ecnt;
19     return ;
20 }
21 void dfs(int u,int pre)
22 {
23     dp[u][0]=dp[u][1]=0;
24     for(int i=head[u];i!=-1;i=edge[i].next)
25     {
26         if(edge[i].to!=pre)
27         {
28             dfs(edge[i].to,u);
29             dp[u][0]+=max(dp[edge[i].to][1],dp[edge[i].to][0]);
30             dp[u][1]+=dp[edge[i].to][0];
31         }
32     }
33     dp[u][1]+=value[u];
34     return ;
35 }
36 int main()
37 {
38     int n,u,v,root,m;
39     while(scanf("%d",&n)!=EOF && n)
40     {
41         for(int i=1;i<=n;i++)
42             scanf("%d",&value[i]);
43         clr_1(head);
44         ecnt=0;
45         clr(fa);
46         while(scanf("%d%d",&u,&v) && u|v)
47         {
48             addedge(v,u);
49             addedge(u,v);
50             fa[u]=v;
51         }
52         for(int i=1;i<=n;i++)
53             if(fa[i]==0)
54             {
55                 root=i;
56                 break;
57             }
58         dfs(root,0);
59         printf("%d\n",max(dp[root][0],dp[root][1]));
60     }
61     return 0;
62 }