[ $\frak{Link}$]

求樹上有P個點的連通塊數目。
求刪除最少多少條邊之後能夠讓樹上出現有P個點的連通塊
考慮點x

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define add_edge(a,b) nxt[++tot]=head[a],head[a]=tot,to[tot]=b
 

int head[155], nxt[310], to[310], siz[155], f[155][155], son[155];
int n, p, tot;
void dfs(int x, int fa) {
    siz[x] = 1;
    for (int i = head[x]; i; i = nxt[i]) {
        if (to[i] == fa) continue;
        dfs(to[i], x);
        siz[x] += siz[to[i]];
        ++son[x];
    }
    f[x][1] = son[x];
    for
 
 (int i = head[x]; i; i = nxt[i]) {
        for (int j = siz[x]; j; --j) {
            for (int k = 0; k <= siz[to[i]]; ++k) {
                if (k > j) continue;
                f[x][j] = min(f[x][j], f[x][j-k] + f[to[i]][k] - 1);
            }
        }
    }
}
int main() {
    memset(f, 0x3f, sizeof(f));
    scanf("%d%d", &n, &p);
    for (int i = 1; i < n; ++i) {
        int a, b;
        scanf("%d%d", &a, &b);
        add_edge(a, b);
        add_edge(b, a);
    }
    dfs(1, 0);
    int ans = f[1][p];
    for (int i = 2; i <= n; ++i) {
        ans = min(ans, f[i][p] + 1);
    }
    printf("%d", ans);
    return 0;
}