160 lines
4.8 KiB
Markdown
160 lines
4.8 KiB
Markdown
##树形$DP$专题
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### 【$T1$】求一颗树的直径
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**难度 ★**
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**[$AcWing$ $1072$. 树的最长路径](https://www.cnblogs.com/littlehb/p/15784687.html)**
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**难度 ★★**
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**[$AcWing$ $1073$. 树的中心](https://www.cnblogs.com/littlehb/p/15786805.html)**
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**[$AcWing$ $1075$. 数字转换](https://www.cnblogs.com/littlehb/p/15787542.html)**
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#### 法一：$dfs$来回两遍
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从任意一点出发找最长$A$，再从$A$找最长；和树形$DP$没有一丁点关系
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**缺点**：不能处理负权边
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```cpp {.line-numbers}
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void dfs1(int u, int sum) {
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st[u] = 1;
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if (sum > ans) {
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ans = sum; // 记录最大距离
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t1 = u; // 记录最远的点t1
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}
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (st[j]) continue;
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dfs1(j, sum + w[i]);
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}
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}
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void dfs2(int u, int sum) {
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st[u] = 1;
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if (sum > ans) {
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ans = sum; // 记录最大距离
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t2 = u; // 记录最远的点t2
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}
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (st[j]) continue;
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dfs2(j, sum + w[i]);
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}
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}
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...
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dfs1(1, 0); // 先找到点距离点1最远的点t
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...
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dfs2(t1, 0); // 找到距离点t最远的点
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```
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#### 法二：树形$DP$
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$d1[u]$表示到$u$的最长路径。$d2[u]$表示到$u$的第二长路径，那么易得 $d1[u]+ d2[u]$即为经过$u$的最长路径
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```cpp {.line-numbers}
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void dfs(int u) {
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st[u] = 1;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (st[j]) continue;
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// 走j子树
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dfs(j);
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// d1[u]:最长路径,d2[u]:次长路径
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if (d1[j] + w[i] >= d1[u])
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d2[u] = d1[u], d1[u] = d1[j] + w[i]; // 最长路转移
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else if (d1[j] + w[i] > d2[u])
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d2[u] = d1[j] + w[i]; // 次长路转移
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}
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// 更新结果
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ans = max(ans, d1[u] + d2[u]);
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}
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```
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###【$T2$】覆盖一棵树的最大、最小权值
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**难度 ★**
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**[$AcWing$ $323$ 战略游戏](https://www.cnblogs.com/littlehb/p/15788291.html)**
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**难度★**
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**[$AcWing$ $285$. 没有上司的舞会](https://www.cnblogs.com/littlehb/p/15468981.html)**
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**对比：**
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没有上司的舞会：每条边上 **最多** 选择一个点，**求最大权值**
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战略游戏：每条边上 **最少** 选择一条点，**求最小权值**
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**难度★★★**
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**[$AcWing$ $1077$. 皇宫看守](https://www.cnblogs.com/littlehb/p/15788813.html)**
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###【$T3$】保留一定数量的点、边，求方案数
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**[$AcWing$ $1074$ . 二叉苹果树 ](https://www.cnblogs.com/littlehb/p/15788100.html)**
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**[计蒜客 $Mila$的苹果树](https://www.jisuanke.com/problem/T3702)** 【**本质是就是二叉苹果树的变形**】
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 2100, M = N << 1;
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const int MOD = 998244353;
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int e[M], h[N], idx, ne[M];
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void add(int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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int n;
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int limit[N]; // 以u为根的子树中数量上限
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int st[N]; // 是不是访问过
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LL f[N][N];
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void dfs(int u) {
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st[u] = 1; // 访问过了
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f[u][0] = 1; // u子树中，放0个苹果的方案是1，此时只能是不管有多少个子结点，都是啥也不能放，方案唯一
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if (limit[u]) f[u][1] = 1; // 在u子树中，如果只放1个的话，那么必然是放在u节点上，否则u子树就不存在的，此时方案数是1
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (st[v]) continue;
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dfs(v);
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// 分组背包
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for (int j = limit[u]; j >= 0; j--) // 枚举u子树体积,倒序降维
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// Q:为什么这里k=0开始是错的呢？
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for (int k = 1; k <= min(j, limit[v]); k++) // 枚举决策,v子树中放资源k个
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f[u][j] = (f[u][j] + f[v][k] * f[u][j - k]) % MOD;
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}
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}
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int main() {
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freopen("appletree.in", "r", stdin);
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freopen("appletree.out", "w", stdout);
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memset(h, -1, sizeof h);
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> limit[i];
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for (int i = 1; i < n; i++) {
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int a, b;
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cin >> a >> b;
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add(a, b), add(b, a); // 无向图
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}
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dfs(1);
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int res = 0;
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for (int i = 0; i <= limit[1]; i++) res = (res + f[1][i]) % MOD;
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printf("%d\n", res);
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return 0;
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}
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```
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**[$51nod$ $1588$ 幸运树](https://www.cnblogs.com/littlehb/p/17203577.html)** 【树形$dp$统计树上方案数】
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**[$P4084$ [$USACO17DEC$]$Barn$ $Painting$ $G$](https://www.cnblogs.com/littlehb/p/17204104.html)**
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**[$P3349$ [$ZJOI2016$]小星星](https://www.luogu.com.cn/problem/P3349)** TODO
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https://blog.csdn.net/A_Bright_CH/article/details/83443279 |