hdu 1520 简单树形dp

G - Anniversary party
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

HDU 1520
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
Sample Output
5

---

题意：题目给出一棵树，每个节点都有其权值。如果选择了一个节点则不可以选择其父节点，问能取得的最大值。
dp[node][0]表示不取该节点
dp[node][1]表示取该节点

#include<iostream>#include<cstdio>#include<algorithm>#include<vector>#include<queue>#include<cmath>#include<cstring>using namespace std;const int maxn = 7100;int in[maxn];int val[maxn];int dp[maxn][2];int vis[maxn];int n, m;vector<int>V[maxn];**void dfs(int u){    vis[u] = true;    dp[u][0] = 0;    dp[u][1] = val[u];    for(unsigned int i=0; i<V[u].size(); ++i)    {        int v = V[u][i];        if(vis[v])            continue;        dfs(v);        dp[u][0] += max(dp[v][1], dp[v][0]);        dp[u][1] += dp[v][0];    }}**int main(){    while(scanf("%d", &n)!=EOF)    {        for(int i=1; i<=n; ++i) V[i].clear();        for(int i=1; i<=n; ++i)            scanf("%d", &val[i]);        memset(in, 0, sizeof(in));        int u, v;        while(scanf("%d%d", &v, &u)&&v+u!=0)        {            V[u].push_back(v);            ++in[v];        }        memset(dp, 0, sizeof(dp));        for(int i=1; i<=n; ++i)if(!in[i])            {                memset(vis, 0, sizeof(vis));                dfs(i);                printf("%d\n", max(dp[i][0], dp[i][1]));                break;            }    }    return 0;}