树形dp

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$简单来说树形dp就是在树上做dp罢了$
$树嘛，就要符合除了根节点外每个节点只有一个父节点$
$然后分析dp的时候，不再是线性的前几的什么属性那种线性术语了$
$而是以i为根节点的树的各种属性$
$状态转移也从原本的前i-1到前i个转移$
$而是u的子树向u为根的树转移$

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上几道例题具体分析
例题一

思路

AC代码

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#include<bits/stdc++.h>

using namespace std;

const int N = 1e4 + 10;

int h[N], e[N], w[N], ne[N], idx;
bool st[N];
int n;
int f[N][3]; //1:最长 0 次长
int res = -0x3f3f3f3f;

void add(int a, int b, int c){
    e[idx] = b;
    ne[idx] = h[a];
    w[idx] = c;
    h[a] = idx ++;
}

void dfs(int u)
{
    for(int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        dfs(j);
        if(f[j][1] + w[i] >= f[u][1]){
            
            f[u][0] = f[u][1];
            f[u][1] = f[j][1] + w[i];
        }
        else if(f[j][1] + w[i] > f[u][0])
        {
            f[u][0] = f[j][1] + w[i];
        }
    }
}

int main()
{
    cin >> n;
    memset(h, -1, sizeof(h));
    for(int i = 1; i <= n - 1; i ++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        st[b] = true;
    }
    int root = 1;
    while(st[root]) root ++;
    dfs(root);
    
    for(int i = 1; i <= n; i ++)
    {
        res = max(res, f[i][0] + f[i][1]);
    }
    
    cout << res << endl;
    
    return 0;
}

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例题2

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;

int h[N], e[N<<1], w[N<<1], ne[N<<1], idx;
int n;
int down1[N], down2[N], up[N], best[N], second[N];

void add(int a, int b, int c)
{
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx ++;
}

void dfs_d(int u, int fa)
{
    for(int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if(j == fa) continue;
        dfs_d(j, u);
        if(down1[j] + w[i] >= down1[u])
        {
            down2[u] = down1[u];
            second[u] = best[u];
            best[u] = j;
            down1[u] = down1[j] + w[i];
        }
        else if(down1[j] + w[i] > down2[u])
        {
            down2[u] = down1[j] + w[i];
            second[u] = j;
        }
    }
}

void dfs_u(int u, int fa)
{
    for(int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if(j == fa) continue;
        if(j == best[u])
        {
            up[j] = max(up[u], down2[u]) + w[i];
        }
        else
        {
            up[j] = max(up[u], down1[u]) + w[i];
        }
        dfs_u(j, u);
    }
}

int main()
{
    memset(h, -1, sizeof(h));
    cin >> n;
    for(int i = 1; i <= n - 1; i ++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    dfs_d(1, -1);
    dfs_u(1, -1);
    int res = 1e9;
    for(int i = 1; i <= n; i ++)
    {
        res = min(max(down1[i], up[i]), res);
    }
    cout << res << endl;
    return 0;
}

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