台州 OJ 2676 Tree of Tree 树状 DP

描述

You're given a tree with weights of each node, you need to find the maximum subtree of specified size of this tree.

Tree Definition
A tree is a connected graph which contains no cycles.

输入

There are several test cases in the input.

The first line of each case are two integers N(1 <= N <= 100), K(1 <= K <= N), where N is the number of nodes of this tree, and K is the subtree's size, followed by a line with N nonnegative integers, where the k-th integer indicates the weight of k-th node. The following N - 1 lines describe the tree, each line are two integers which means there is an edge between these two nodes. All indices above are zero-base and it is guaranteed that the description of the tree is correct.

输出

One line with a single integer for each case, which is the total weights of the maximum subtree.

给一棵树（都是无向边，所以根的位置可以是任意的），求指定大小子树的最大重量和。

第一次自己做出树状DP题目，好开心~

dp[u][j] 表示，以 u 为根的子树，大小为 j 时，最大的重量和

则可以写出状态转移方程 dp[u][j] = max(dp[u][k] + dp[v][j-k]) v 为 u 的子树，为当前树大小为 k，子树大小为 j-k，因为根结点是一定要地，所以 1<k<=j。

还是要注意，遍历树的大小的那个循环次序，是从大到小，类似于 0-1 背包的一维数组优化一样，要搞清楚数组中每一个格子存的状态到底是什么。

代码：

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

const int MAX = 105; 

struct Node{
    int num, weight;
    Node(){}
    Node(int nn, int nw):num(nn), weight(nw){}
}; 

vector<int> tree[MAX];
int vis[MAX];
int dp[MAX][MAX];    //在 i 结点，结点数量为 j 时的最大重量 
int w[MAX];
int n, k;
int ans;

int dfs(int u);

int main(){
//    freopen("input.txt", "r", stdin);
    
    while(cin >> n >> k){
        for(int i=0; i<n; i++){
            cin >> w[i];
            tree[i].clear();
        }
        int u, v;
        for(int i=1; i<n; i++){
            cin >> u >> v;
            tree[u].push_back(v);
            tree[v].push_back(u);
        }
        
        ans = 0;
        for(int i=0; i<n; i++){
            memset(vis, 0, sizeof(vis));
            memset(dp, -1, sizeof(dp));
            for(int j=0; j<n; j++){
                dp[j][0] = 0;
                dp[j][1] = w[j];
            }
            dfs(i);
        }
        
        cout << ans << endl;
    }
    
    return 0;
}

int dfs(int u){
//    cout << u << endl;
    vis[u] = 1;
    int maxNum = 1;
    for(int i=0; i<tree[u].size(); i++){
        int v = tree[u][i];
        if(vis[v] == 1)
            continue;
        maxNum += dfs(v);
        for(int j=min(maxNum, k); j>=1; j--){        //一共 j 个 ,注意循环次序，从大到小，前面的状态要搞懂 
            for(int x=0; x<j; x++){        //子树放 x 个 （u结点要放，因此 x 最大为 j-1） 
                if(dp[u][j-x] != -1 && dp[v][x] != -1)
                    dp[u][j] = max(dp[u][j], dp[u][j-x] + dp[v][x]);
            }
        }
    }
    
    ans = max(ans, dp[u][k]);
    return maxNum;
//    cout << u << " " << k << " " << dp[u][k] << endl;
}