树形dp入门（poj 2342 Anniversary party）
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。
解题思路：
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
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
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
typedef long long ll;
const int maxn=6005;
const int INF=0x3f3f3f3f;
//先建树,找到根节点,然后开始dfs
int n,m,x,y;
int dp[maxn][2],vis[maxn],father[maxn];//dp[i][1]表示来,dp[i][0]表示不来
void dfs(int node)
{
int i,j;
vis[node]=1;
for(i=1; i<=n; i++)
{
if(!vis[i]&&father[i]==node)
{
dfs(i);//从根节点的儿子节点开始递归
dp[node][1]+=dp[i][0];//上司来
dp[node][0]+=max(dp[i][1],dp[i][0]);//上司不来,然后看下属来还是不来
}
}
}
int main()
{
while(~scanf("%d",&n))
{
int i,j;
memset(father,0,sizeof(father));
memset(dp,0,sizeof(dp));
memset(vis,0,sizeof(vis));
for(i=1; i<=n; i++)
scanf("%d",&dp[i][1]);
int root=0;
//bool flag=1;
while(~scanf("%d %d",&x,&y)&&(x!=0||y!=0))
{
father[x]=y;
//if(root==x||flag)
//root=y;
}
root=y;
while(father[root])
root=father[root];//找根节点
dfs(root);
int maxx=max(dp[root][1],dp[root][0]);
printf("%d\n",maxx);
}
return 0;
}
//更快的一种方法
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
//typedef long long ll;
//const int maxn=6005;
//const int INF=0x3f3f3f3f;
struct node
{
int child,father,brother,present,not_present;
int max()//结构体内的函数调用时耗时少
{
return present>not_present?present:not_present;
}
void init()
{
child=father=brother=not_present=0;
}
} tree[6005];
void dfs(int root)
{
int son = tree[root].child;
while(son)
{
dfs(son);
tree[root].present+=tree[son].not_present;
tree[root].not_present+=tree[son].max();
son = tree[son].brother;
}
}
int main()
{
int n,i,j,k,l;
while(~scanf("%d",&n)&&n)
{
for(i = 1; i<=n; i++)
{
scanf("%d",&tree[i].present);
tree[i].init();
}
while(~scanf("%d%d",&l,&k),l+k)
{
tree[l].father = k;
tree[l].brother = tree[k].child;
tree[k].child = l;
}
for(i = 1; i<=n; i++)
{
if(!tree[i].father)
{
dfs(i);
printf("%d\n",tree[i].max());
break;
}
}
}
return 0;
}
