DP Intro - Tree POJ2342 Anniversary party

POJ 2342 Anniversary party (树形dp 入门题)

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4810		Accepted: 2724

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

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

Ural State University Internal Contest October'2000 Students Session

题目链接：http://poj.org/problem?id=2342

题目大意：一棵树，每个节点有一个值，现在要从中选一些点，要求这些点值和最大并且每对儿子和父亲不能同时被选

题目分析：dp[i][0]和dp[i][1]分别表示不选和选第i个点时以i为子树根时子树值的和，则：
dp[fa[i]][1] += dp[i][0] 表示选i的父亲时，其值等于自身值加上不选i时的值
dp[fa[i]][0] += max(dp[i][1], dp[i][0]) 表示不选父亲时，则其值等于其儿子被选或没被选的值的最大值

#include
#include
#include
#include
using namespace std;
int const MAX = 6005;
int dp[MAX][2], val[MAX];
bool vis[MAX], flag[MAX];
vector  vt[MAX];

void DFS(int fa)
{
    vis[fa] = true;
    dp[fa][1] = val[fa];
    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][1], dp[son][0]);
        }
    }
    return;
}

int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int i = 1; i <= n; i++)
            vt[i].clear();
        memset(flag, false, sizeof(flag));
        memset(vis, false, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d", &val[i]);
        for(int i = 0; i < n - 1; i++)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            vt[b].push_back(a);
            flag[a] = true;
        }
        int root;
        for(int i = 1; i <= n; i++) //n个点n－1条边，必然存在“入度”为0的点即树根
        {
            if(!flag[i])
            {
                root = i;
                break;
            }
        }
        DFS(root);
        printf("%d\n", max(dp[root][1], dp[root][0]));
    }
}

=============================================================================================

今天开始做老师给的专辑，打开DP专辑 A题 Rebuilding Roads 直接不会了，发现是树形DP，百度了下了该题，看了老半天看不懂，想死的冲动都有了~~~~

最后百度了下，树形DP入门，找到了 poj 2342 Anniversary party   先入门一下~

题意：

某公司要举办一次晚会，但是为了使得晚会的气氛更加活跃，每个参加晚会的人都不希望在晚会中见到他的直接上司，现在已知每个人的活跃指数和上司关系（当然不可能存在环），求邀请哪些人（多少人）来能使得晚会的总活跃指数最大。

思路：

任何一个点的取舍可以看作一种决策，那么状态就是在某个点取的时候或者不取的时候，以他为根的子树能有的最大活跃总值。分别可以用f[i,1]和f[i,0]表示第i个人来和不来。

当i来的时候，dp[i][1] += dp[j][0];//j为i的下属

当i不来的时候，dp[i][0] +=max(dp[j][1],dp[j][0]);//j为i的下属

以下代码参考：http://hi.baidu.com/saintlleo/blog/item/0606b3feb7026ad3b48f3111.html

//AC CODE:

1. #include<iostream>
2. #include<cmath>
3. #include<algorithm>
4. #include<vector>
5. #include<cstdio>
6. #include<cstdlib>
7. #include<cstring>
8. #include<string>
9. using namespace std;
10. #define maxn 6005
11. int n;
12. int dp[maxn][2],father[maxn];//dp[i][0]0表示不去，dp[i][1]1表示去了
13. bool visited[maxn];
14. void tree_dp(int node)
15. {
16. int i;
17. visited[node] = 1;
18. for(i=1; i<=n; i++)
19. {
20. if(!visited[i]&&father[i] == node)//i为下属
21. {
22. tree_dp(i);//递归调用孩子结点，从叶子结点开始dp
23. //关键
24. dp[node][1] += dp[i][0];//上司来,下属不来
25. dp[node][0] +=max(dp[i][1],dp[i][0]);//上司不来，下属来、不来
26. }
27. }
28. }
29. int main()
30. {
31. int i;
32. int f,c,root;
33. while(scanf("%d",&n)!=EOF)
34. {
35. memset(dp,0,sizeof(dp));
36. memset(father,0,sizeof(father));
37. memset(visited,0,sizeof(visited));
38. for(i=1; i<=n; i++)
39. {
40. scanf("%d",&dp[i][1]);
41. }
42. root = 0;//记录父结点
43. bool beg = 1;
44. while (scanf("%d %d",&c,&f),c||f)
45. {
46. father[c] = f;
47. if( root == c || beg )
48. {
49. root = f;
50. }
51. }
52. while(father[root])//查找父结点
53. root=father[root];
54. tree_dp(root);
55. int imax=max(dp[root][0],dp[root][1]);
56. printf("%d\n",imax);
57. }
58. return 0;
59. }