Problem

给你一棵树，最少删掉哪些边，能使得余下的至少有1个大小刚好为k的残树。

1 ≤ k ≤ n ≤ 400

Solution

用f[i][j]表示以i为根有j个节点的最少删边数量

因为此题要输出删除的边

v[i][j]表示以i为根删掉j个节点需要删去的边对应的(点u，该u点还需要删去的边数量)

Notice

因为要输出方案，所以比较复杂

Code

#include<cmath>

#include<vector>

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

#define sqz main

#define ll long long

#define reg register int

#define rep(i, a, b) for (reg i = a; i <= b; i++)

#define per(i, a, b) for (reg i = a; i >= b; i--)

#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)

typedef pair<int, int> Node;

const int INF = 1e9, N = 400;

const double eps = 1e-6, phi = acos(-1.0);

ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}

ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();

if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}

void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}

struct node

{

	int vet, next;

}edge[N + 5];

int head[N + 5], num = 0, f[N + 5][N + 5], Flag[N + 5], fa[N + 5], n, k;

vector<Node> v[N + 5][N + 5];

void add(int u, int v)

{

	edge[++num].vet = v;

	edge[num].next = head[u];

	head[u] = num;

}

void dfs(int u)

{

    rep(i, 0, k) f[u][i] = INF;

    f[u][1] = 0;

	travel(i, u)

	{

		int vv = edge[i].vet;

		if (vv == fa[u]) continue;

		fa[vv] = u;

		dfs(vv);

		per(j, k, 0)

		{

			f[u][j]++;

			rep(t, 0, j)

                if (f[u][j - t] + f[vv][t] < f[u][j])

                {

                    f[u][j] = f[u][j - t] + f[vv][t];

                    v[u][j].clear();

                    v[u][j].assign(v[u][j - t].begin(), v[u][j - t].end());

                    v[u][j].push_back(make_pair(vv, t));

                }

		}

	}

}

void Out(int u, int now)

{

    Flag[u] = 1;

    for (auto i : v[u][now]) Out(i.first, i.second);

    travel(i, u)

        if (!Flag[edge[i].vet]) printf("%d ", (i + 1) / 2);

}

int sqz()

{

	n = read(), k = read();

	rep(i, 1, n) head[i] = 0;

	rep(i, 1, n - 1)

	{

		int u = read(), v = read();

		add(u, v), add(v, u);

	}

	dfs(1);

	int ans = f[1][k], root = 1;

	rep(i, 2, n)

		if (f[i][k] + 1 < ans) ans = f[i][k] + 1, root = i;

	printf("%d\n", ans) ;

	if (ans) Out(root, k);

	puts("");

	return 0;

}