Luo Gu 1273: cable television network

Luo Gu 1273: cable television network

Thinking

• Tree grouping backpack.
• Set \ (f (i, j) \) represented by \ (i \) subtree rooted in satisfying \ (j \) the maximum benefit available to customers.
• Then the last find the most customers when the reverse cycle, find the first is greater than \ (0 \ leq f (root , j) \) The \ (j \) case, which is able to meet most users do not lose money.
• Transfer equation: \ (F (I, J) = max (F (I, J), F (I, JK) + F (Son, K) -cos) \) .
• Where \ (cos \) represents the cost of this edge.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3000 + 10;

int n, m; //节点数 叶子节点数
int a[maxn];
int head[maxn<<1], edge[maxn<<1], ver[maxn<<1], nex[maxn<<1], tot;
inline void add_edge(int x, int y, int z){
    ver[++tot] = y; edge[tot] = z;
    nex[tot] = head[x]; head[x] = tot;
}

int f[maxn][maxn];

int dfs(int x)
{
    if(x > n - m)
    {
        f[x][1] = a[x];
        return 1;
    }
    int sum = 0, t; //sum表示该节点最多选多少个用户，也就是背包的容量
    for(int i = head[x]; i; i = nex[i])
    {
        int y = ver[i], z = edge[i];
        t = dfs(y); sum += t; 
        for(int j = sum; j >= 0; j--)
            for(int k = 0; k <= t; k++) //t是子树的用户端 枚举选1个，2个....
                if(j - k >= 0)
                f[x][j] = max(f[x][j], f[x][j-k]+f[y][k]-z);
    } return sum;
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1, k, y, z; i <= n-m; i++)
    {
        scanf("%d", &k);
        for(int j = 1; j <= k; j++)
        {
            scanf("%d%d", &y, &z);
            add_edge(i, y, z);
        }
    }
    for(int i = n - m + 1; i <= n; i++)
        scanf("%d", &a[i]);
    memset(f, 0xcf, sizeof f);
    //什么都不选的价值就是0
    for(int i = 1; i <= n; i++)
        f[i][0] = 0; 
    dfs(1);
    for(int i = m; i >= 1; i--)
        if(f[1][i] >= 0)
        {
            cout << i << endl;
            break;
        }
    return 0;
}