poj 1155 TELE

树形dp一道，开始想到的状态table[i][j]表示节点i为根的子树花费上限为j所能供给的最多观众,后来发现花费并没有办法得出，所以换另一种状态表示table[i][j]表示以节点i为根的子树供给j位观众所能得到的最大利益，

状态转移方程为 table[i][j] = max(∑table[a][b])(a为i的儿子节点,∑b = j)，复杂度为O(n*m^2),貌似要超时，不过数据还是没那么恐怖的，没有想到怎么优化到O(n*m)。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits> 
#include <set>
#include <string> 
#include <sstream>
#include <utility>
#include <ctime>
//#pragma comment(linker, "/STACK:1024000000,1024000000") 

using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::getline;
using std::make_pair;
using std::greater;

const int MAXN(3010);
const int INFI((INT_MAX-1) >> 1);

struct EDGE
{
	int v, w;
	EDGE *next;
};

EDGE *first[MAXN];
EDGE pool[MAXN];
EDGE *rear;


void init()
{
	memset(first, 0, sizeof(first));
	rear = pool;
}

void insert(int tu, int tv, int tw)
{
	rear->v = tv;
	rear->w = tw;
	rear->next = first[tu];
	first[tu] = rear++;
}

int table[MAXN][MAXN];
int value[MAXN];
int num[MAXN];
int N, M, tm;



void dfs(int cur, int limit, int tw)
{
	if(cur >= tm)
	{
		table[cur][0] = -tw;
		table[cur][1] = value[cur]-tw;
		num[cur] = 1;
		return;
	}
	table[cur][0] = -tw;
	int temp = 0;
	for(EDGE *i = first[cur]; i; i = i->next)
	{
		dfs(i->v, limit, i->w);
		temp += num[i->v];
		for(int j = temp; j >= 0; --j)
			for(int k = 0; k <= j; ++k)
				table[cur][j] = max(table[cur][j], table[cur][j-k]+table[i->v][k]);
	}
	num[cur] = temp;
}

int main()
{
	while(~scanf("%d%d", &N, &M))
	{
		init();
		tm = N-M+1;
		int tv, tw;
		for(int i = 1; i < tm; ++i)
		{
			int tl;
			scanf("%d", &tl);
			for(int j = 0; j < tl; ++j)
			{
				scanf("%d%d", &tv, &tw);
				insert(i, tv, tw);
			}
		}
		for(int i = tm; i <= N; ++i)
		{
			scanf("%d", value+i);
		}
		for(int i = 1; i <= N; ++i)
		{
			for(int j = 0; j <= M; ++j)
				table[i][j] = -INFI;
			num[i] = 0;
		}
		dfs(1, M, 0);
		int ans = M;
		while(table[1][ans] < 0)
			--ans;
		printf("%d\n", ans);
	}
	return 0;
}