POJ 2342 Anniversary party （树形dp入门）

Anniversary party

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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

711111111 32 36 47 44 53 50 0 

 

Sample Output

5 题意：一个party中有上下级关系的人不能同时参加，求最大愉悦度树形dp入门，如果一个人参加，那么取他下属不参加的最大值如果不参加，取他下属参加或者不参加的最大值转移方程： dp[k][1]+=dp[v[i]][0];dp[k][0]+=max(dp[v[i]][0],dp[v[i]][1]);
#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3f#define maxn 12005int dp[maxn][2],val[maxn],in[maxn];int fir[maxn],nex[maxn],u[maxn],v[maxn];int e_max;void init(){    memset(in,0,sizeof in);    memset(dp,0,sizeof dp);    memset(fir,-1,sizeof fir);    e_max=0;}void add_edge(int s,int t){    int e=e_max++;    u[e]=s;    v[e]=t;    nex[e]=fir[s];    fir[s]=e;}int dfs(int k){    for(int i=fir[k]; ~i; i=nex[i])    {        dfs(v[i]);        dp[k][1]+=dp[v[i]][0];        dp[k][0]+=max(dp[v[i]][0],dp[v[i]][1]);    }}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        init();        int root=0;        for(int i=1; i<=n; i++)            scanf("%d",&val[i]),dp[i][1]=val[i];        int l,k;        while(scanf("%d%d",&l,&k)!=EOF&&l+k!=0)        {            add_edge(k,l);            in[l]++;        }        for(int i=1;i<=n;i++)            if(in[i]==0) root=i;        dfs(root);        printf("%d\n",max(dp[root][0],dp[root][1]));    }    return 0;}