【题目链接】

点击打开链接

【算法】

f[i][j]表示以i为根的子树中,最少删多少条边可以组成j个节点的子树

树上背包,即可

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 200
const int INF = 1e8; int i,j,n,p,x,y,ans = INF;
int sum[MAXN],f[MAXN][MAXN];
vector<int> e[MAXN]; inline void dfs(int x)
{
int i,j,k,y;
sum[x] = ;
for (i = ; i < e[x].size(); i++)
{
y = e[x][i];
dfs(y);
sum[x] += sum[y];
for (j = sum[x]; j > ; j--)
{
for (k = ; k < j; k++)
{
f[x][j] = min(f[x][j],f[x][j-k]+f[y][k]-);
}
}
}
} int main()
{ scanf("%d%d",&n,&p);
for (i = ; i <= n; i++)
{
for (j = ; j <= n; j++)
{
if (j > ) f[i][j] = INF;
}
}
for (i = ; i < n; i++)
{
scanf("%d%d",&x,&y);
e[x].push_back(y);
f[x][]++;
}
dfs();
for (i = ; i <= n; i++)
{
if (i == ) ans = min(ans,f[i][p]);
else ans = min(ans,f[i][p]+);
}
printf("%d\n",ans); return ; }

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