Codeforces 440 D. Berland Federalization 树形DP，记录DP

题目链接：http://codeforces.com/contest/440/problem/D

Recently, Berland faces federalization requests more and more often. The proponents propose to divide the country into separate states. Moreover, they demand that there is a state which includes exactly k towns.

Currently, Berland has n towns, some pairs of them are connected by bilateral roads. Berland has only n - 1 roads. You can reach any city from the capital, that is, the road network forms a tree.

The Ministry of Roads fears that after the reform those roads that will connect the towns of different states will bring a lot of trouble.

Your task is to come up with a plan to divide the country into states such that:

each state is connected, i.e. for each state it is possible to get from any town to any other using its roads (that is, the roads that connect the state towns),
there is a state that consisted of exactly k cities,
the number of roads that connect different states is minimum.

Input

The first line contains integers n, k (1 ≤ k ≤ n ≤ 400). Then follow n - 1 lines, each of them describes a road in Berland. The roads are given as pairs of integers xi, yi (1 ≤ xi, yi ≤ n; xi ≠ yi) — the numbers of towns connected by the road. Assume that the towns are numbered from 1 to n.

Output

The the first line print the required minimum number of "problem" roads t. Then print a sequence of t integers — their indices in the found division. The roads are numbered starting from 1 in the order they follow in the input. If there are multiple possible solutions, print any of them.

If the solution shows that there are no "problem" roads at all, print a single integer 0 and either leave the second line empty or do not print it at all.

题意：

　　给你一棵树，最少的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

　　给出方案

题解：

　　DP[i][j]表示以i为根删掉j个节点的最少删边数量

　　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)

　　转移较简单

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e2+10, M = 1e3+20, mod = 1e9+7, inf = 2e9;

int dp[N][N],sz[N],n,K,a,b,fa[N];
vector<pair<int,int >  > G[N];
vector< pair<int ,int > > V[N][N];
void dfs(int u,int f) {
    sz[u] = 1;
    for(int i = 0; i < G[u].size(); ++i) {
        int to = G[u][i].first;
        if(to == f) continue;
        dfs(to,u);
        fa[to] = G[u][i].second;
        sz[u] += sz[to];
    }
}
void dfs2(int u,int f,int index) {
    for(int i = 0; i < G[u].size(); ++i) {
        int to = G[u][i].first;
        if(to == f) continue;
        dfs2(to,u,G[u][i].second);
        for(int j = sz[u]; j >= 0; --j)
            for(int k = 0; k <= min(sz[to],j); ++k)
            {
                if(dp[u][j - k] + dp[to][k] < dp[u][j]) {
                     dp[u][j] = dp[u][j - k] + dp[to][k];
                     V[u][j] = V[u][j-k];
                     V[u][j].push_back(MP(to,k));
                }
            }
    }
    if(f) {
        dp[u][sz[u]] = 1;
        V[u][sz[u]].clear();
        V[u][sz[u]].push_back(MP(u,sz[u]));
    }
    else dp[u][sz[u]] = 0;
}
int vis[N];
void work(int u,int x) {
    vis[u] = 1;
    if(sz[u] == x) {
        cout<<fa[u]<<" ";
        return ;
    }
    for(int i = 0; i < V[u][x].size(); ++i) {
        work(V[u][x][i].first,V[u][x][i].second);
    }
}
int main() {
    scanf("%d%d",&n,&K);
    for(int i = 1; i < n; ++i) {
        scanf("%d%d",&a,&b);
        G[a].push_back(MP(b,i));
        G[b].push_back(MP(a,i));
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= n; ++j) dp[i][j] = inf;
    }
    dfs(1,0);
    dfs2(1,0,0);
    int ans = dp[1][n-K],ansj = 1;
    for(int i = 2; i <= n; ++i) {
        if(sz[i] >= K && 1+dp[i][sz[i]-K] < ans) {
            ans = min(ans,1+dp[i][sz[i]-K]);
            ansj = i;
        }
    }
    cout<<ans<<endl;
    if(ansj != 1) cout<<fa[ansj]<<" ";
    work(ansj,sz[ansj] - K);
    return 0;
}