dp [root] [j]: the root of the subtree root, to obtain the number of sides of the sub-tree nodes j need to lose the least, attention must be retained in the sub-tree root node. Otherwise it is impossible dp
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<vector> using namespace std; inline int read(){ int sum=0,x=1; char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-') x=0; ch=getchar(); } while(ch>='0'&&ch<='9'){ sum=(sum<<1)+(sum<<3)+(ch^48),ch=getchar(); } return x?sum:-sum; } inline void write(int x){ if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10+'0'); } int mi(int x,int y){ return x<y?x:y; } int ma(int x,int y){ return x>y?x:y; } const int M=155; const int inf =0x3f3f3f3f; vector<int>g[M]; int dp[M][M],num[M],sz[M]; void dfs(int u){ sz[u]=1; if(g[u].size()==0){ dp[u][1]=0; sz[u]=1; return ; } for(int i=0;i<g[u].size();i++){ int v=g[u][i]; dfs(v); sz[u]+=sz[v]; for(int j=sz[u];j>=0;j--){ for(int k=1;k<=j;k++){ dp[u][j]=mi(dp[u][j],dp[u][j-k]+dp[v][k]-1); } } } } int main(){ int n=read(),p=read(); for(int i=0;i<=n;i++) for(int j=0;j<=p;j++) dp[i][j]=inf; for(int i=1;i<n;i++){ int x,y; x=read(); y=read(); g[x].push_back(y); num[x]++; } for(int i=1;i<=n;i++) dp[i][1]=num[i]; dfs(1); int ans=dp[1][p]; for(int i=2;i<=n;i++) ans=mi(ans,dp[i][p]+1); write(ans); putchar('\n'); return 0; }