TopcoderSRM679 Div1 250 FiringEmployees(树形dp)

题意

[题目链接]这怎么发链接啊。。。。。

有一个 \(n\) 个点的树，每个点有点权（点权可能为负） ，求包含点\(1\)的最
大权连通子图（的权值和） 。
\(n \leqslant 2500\)

Sol

刚开始还以为是个树形依赖背包呢。。结果发现后面给的两个vector根本就没用

直接减一下得到每个点的点权，然后xjb dp一波

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int a[MAXN], f[MAXN];
vector<int> v[MAXN];
class FiringEmployees{
public:
    void dfs(int x, int fa) {
        f[x] = a[x];
        for(int i = 0, to; i < v[x].size(); i++) {
            if((to = v[x][i]) == fa) continue;
            dfs(to, x);
            f[x] = max(f[x], f[x] + f[to]);
        }
    }
    int fire(vector <int> fa, vector <int> salary, vector <int> productivity) {
        int N = fa.size();
        for(int i = 1; i <= N; i++) {
            a[i] = productivity[i - 1] - salary[i - 1];
            //cout << fa[i - 1] << endl;
            v[fa[i - 1]].push_back(i);

        }
        dfs(0, -1);
        return f[0];
    }
};

int main() {
    int N = read();
    vector<int> a, b, c;
    for(int i = 1; i <= N; i++) a.push_back(read());
    for(int i = 1; i <= N; i++) b.push_back(read());
    for(int i = 1; i <= N; i++) c.push_back(read());
    cout << FiringEmployees().fire(a, b, c);
}
/*
6
0 0 1 1 2 2
1 1 1 2 2 2
2 2 2 1 1 1

9
0 1 2 1 2 3 4 2 3
5 3 6 8 4 2 4 6 7
2 5 7 8 5 3 5 7 9


2
0 1
1 10
5 5

4
{{0, 1, 2, 3}
{4, 3, 2, 1}
{2, 3, 4, 5}}
*/