CF1324 --- Maximum White Subtree

题干

You are given a tree consisting of $n$ vertices. A tree is a connected undirected graph with $n−1$ edges. Each vertex $v$ of this tree has a color assigned to it ($a_v$=1 if the vertex $v$ is white and $0$ if the vertex $v$ is black).

You have to solve the following problem for each vertex $v$: what is the maximum difference between the number of white and the number of black vertices you >can obtain if you choose some subtree of the given tree that contains the vertex $v$? The subtree of the tree is the connected subgraph of the given tree. More >formally, if you choose the subtree that contains $cnt_w$ white vertices and $cnt_b$ black vertices, you have to maximize $cnt_w−cnt_b$.

$\mathcal{Input}$

The first line of the input contains one integer $n$ ($2 \leq n \leq 2 * 10^5$) — the number of vertices in the tree.

The second line of the input contains $n$ integers $a_1,a_2,\cdots, a_n (0\leq a_i \leq 1)$, where $a_i$ is the color of the $i-th$ vertex.

Each of the next $n−1$ lines describes an edge of the tree. Edge $i$ is denoted by two integers $u_i$ and $v_i$, the labels of vertices it connects $(1 \leq u_i,v_i \leq n,u_i \not= v_i)$.

It is guaranteed that the given edges form a tree.

$\mathcal{Output}$

Print $n$ integers $res_1, res_2, \cdots, res_n$, where $res_i$ is the maximum possible difference between the number of white and black vertices in some subtree that contains the vertex $i$.

$\mathcal{Example}$

$Case_1$

$Input$

9

0 1 1 1 0 0 0 0 1

1 2

1 3

3 4

3 5

2 6

4 7

6 8

5 9

$Output$

2 2 2 2 2 1 1 0 2

$Case_2$

$Input$

4

0 0 1 0

1 2

1 3

1 4

$Output$

0 -1 1 -1

$\mathcal{Note}$

The first example is shown below:

The black vertices have bold borders.

In the second example, the best subtree for vertices $2$,$3$ and $4$ are vertices $2$,$3$ and $4$ correspondingly. And the best subtree for the vertex $1$ is the subtree consisting of vertices $1$ and $3$.

$\mathcal{Tag}$

dfs and similar dp *1800

思路分析

这题要求的是求出对任何一个vertex $v$，求出包含这个节点的子树$cnt_w - cnt_b$的最大值。

暴力想法

首先思考下暴力写法应该如何写。

所以，对于所有可能的路径的贡献值的累加，且贡献值需大于等于$0$.不妨设$dp[v]$代表该结点的最大值。故

\[dp[v] = a_v + \sum_{i=0}^{k}max(0, dp[i])
\]

假如用暴力写法，就是对于每个结点$v$，暴力搜索所有的adjacent结点，利用$dfs$暴力搜索。但是结点最大为$2*10^5$这个暴力算法显然会超时，考虑如何优化。

算法优化

对于从下往上的贡献，可以利用从下往上的$dfs$树形$dp$进行获取,剩余的就是刨去以$v$为根的子树的贡献值比较难求。在这里我们设$fa$为结点$v$的父节点。$f_v$代表从下往上以$v$为根的白点数减去黑点数的最大值.$dp[v]$代表最终的最大值。因此根据刨去以$v$为根的子树的贡献值这个思想，我们可以发现:

\[add = dp[fa] - max(0, f_v)
\]

就是刨去以$v$为根的子树的贡献值。因此最终我们可以写出状态转移方程：

\[ dp[v] =\left\{\begin{array}[ll]
1f[v] & if \;v = root \\
f[v] + max(0, dp[fa] - max(0, f[v])) & if\; v \not= root\\
\end{array}\right.
\]

因此最后我们的思路为:

从下往上树形$dp$,计算$f_v$
从上往下换根$dp$,计算$dp[v]$

代码

#include<bits/stdc++.h>

using namespace std;

using VI = vector<int>;

using VVI = vector<VI>;

VI a;

VI dp;

VI ans;

VVI e;

void dfs(int x, int fa = -1){

    dp[x] = a[x];

    for (int to : e[x]){

        if (to == fa) continue;

        dfs(to, x);

        dp[x] += max(0, dp[to]);

    }

}

void rdfs(int x, int fa = -1){

    ans[x] = dp[x];

    for (int to : e[x]){

        if (to == fa) continue;

        dp[x] -= max(0, dp[to]);

        dp[to] += max(0, dp[x]);

        rdfs(to, x);

        dp[to] -= max(0, dp[x]) ;

        dp[x] += max(0, dp[to]);

    }

}

int main(){

    int n; cin >> n;

    a = dp = ans = VI(n);

    e = VVI(n);

    for (int i = 0; i < n; ++ i){

            cin >> a[i];

            if (a[i] == 0) a[i] = -1;

    }

    for (int i = 0; i < n - 1; ++ i){

        int x, y;

        cin >> x >> y;

        -- x, -- y;

        e[x].push_back(y);

        e[y].push_back(x);

    }

    dfs(0);

    rdfs(0);

    for (int ret : ans) cout << ret << " ";

    cout << endl;

    return 0;

}

第一次在博客园写题解，加油加油！每天一个CF题！