题目链接：
题意：给出一棵树，找出一个不大于长度为m的链，使得其他点到该链的长度之和最小。
思路：首先，链有两种。对于一个子树u，或者u一直向下成竖线形状，或者以u为转折点以及u的两个孩子组成的形状。首先预处理出以下数组：
（1）downSum[u]:u的所有孩子到达u的距离，downCnt[u]：u子树节点个数；
（2）upSum[u]:除u子树的节点外其他节点到达u的距离；upCnt[u]:除u子树的节点个数。
树形DP时，竖线形状的比较简单，我们用dp[u][L]表示以u为子树，向下有一条长度为L的链的最小代价；然后dp[u][m]+upSum[u]就是答案；对于折线形状的，设以u的两个孩子v1、v2组成，则答案为upSum[u]+downSum[u]+dp[v1][j-1]-downSum[v1]-downCnt[v1]+dp[v2][m-1-j]-downSum[v2]-downCnt[v2]，因此只要维护upSum[u]+downSum[u]+dp[v1][j-1]-downSum[v1]-downCnt[v1]的最小值即可。
#include <iostream>
#include <cstdio>#include <string.h>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include <stack>#include <string>#include <map>#include <ctype.h>#include <time.h> #define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define EPS 1e-10 #define FOR0(i,x) for(i=0;i<x;i++)#define FOR1(i,x) for(i=1;i<=x;i++)#define FOR(i,a,b) for(i=a;i<=b;i++)#define FORL0(i,a) for(i=a;i>=0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--) #define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std; void RD(int &x){scanf("%d",&x);}void RD(i64 &x){scanf("%I64d",&x);}void RD(u64 &x){scanf("%I64u",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(i64 &x,i64 &y){scanf("%I64d%I64d",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%I64d%I64d%I64d",&x,&y,&z);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;} void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(i64 x) {printf("%lld\n",x);}void PR(i64 x,i64 y) {printf("%I64d %I64d\n",x,y);}void PR(u32 x) {printf("%u\n",x);}void PR(u64 x) {printf("%I64u\n",x);}void PR(double x) {printf("%.6lf\n",x);}void PR(double x,double y) {printf("%.5lf %.5lf\n",x,y);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout<<x<<endl;} void upMin(int &x,int y) {if(x>y) x=y;}void upMin(i64 &x,i64 y) {if(x>y) x=y;}void upMin(double &x,double y) {if(x>y) x=y;}void upMax(int &x,int y) {if(x<y) x=y;}void upMax(i64 &x,i64 y) {if(x<y) x=y;}void upMax(double &x,double y) {if(x<y) x=y;} const int mod=10000;const double dinf=1e12;const i64 inf=((i64)1)<<60;const int INF=1000000000;const int N=10005;int n,m;vector<int> g[N];int downSum[N],upSum[N],downCnt[N],upCnt[N];void DFS1(int u,int pre){ downCnt[u]=1; downSum[u]=0; int i,v; FOR0(i,SZ(g[u])) { v=g[u][i]; if(v==pre) continue; DFS1(v,u); downCnt[u]+=downCnt[v]; downSum[u]+=downSum[v]+downCnt[v]; }}void DFS2(int u,int pre){ if(pre!=-1) { upCnt[u]=n-downCnt[u]; upSum[u]=upSum[pre]+upCnt[pre]+downSum[pre]-downSum[u]-(downCnt[pre]-1)+(downCnt[pre]-1-downCnt[u])*2+1; } else upCnt[u]=upSum[u]=0; int i,v; FOR0(i,SZ(g[u])) { v=g[u][i]; if(v==pre) continue; DFS2(v,u); }}int dp[N][105],f[N][105];int ans;void DFS(int u,int pre){ int i,j,v; FOR0(i,m+1) dp[u][i]=downSum[u],f[u][i]=downSum[u]+upSum[u]; FOR0(i,SZ(g[u])) { v=g[u][i]; if(v==pre) continue; DFS(v,u); for(j=1;j<=m;j++) { upMin(dp[u][j],dp[v][j-1]+downSum[u]-downSum[v]-downCnt[v]); if(m-1-j>=0) upMin(ans,f[u][j]+dp[v][m-1-j]-downSum[v]-downCnt[v]); } for(j=1;j<=m;j++) upMin(f[u][j],upSum[u]+downSum[u]+dp[v][j-1]-downSum[v]-downCnt[v]); } upMin(ans,dp[u][m]+upSum[u]);}int main(){ Rush(n) { RD(m); if(n==0&&m==0) break; int i; FOR0(i,n) g[i].clear(); int x,y; FOR1(i,n-1) { RD(x,y); g[x].pb(y); g[y].pb(x); } DFS1(0,-1); DFS2(0,-1); ans=INF; DFS(0,-1); PR(ans); }}