【樹狀DP】ural 1018

還有些不懂，先放下

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}#define LL long long#define PI acos(-1.0)#define N  401#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)int max(int a,int b){return a>b?a:b;}struct node{    int l,r;    int val;}t[110];int g[110][110];int n,m;int ans[110][110];bool vis[110];void Build(int p){//p=1    int i,j,k;    bool tag=1;    vis[p]=1;    for(i=1;i<=n;i++){        if(g[p][i]!=-1 && !vis[i]){            if(tag){                tag=0;                t[p].l=i;                t[i].val=g[p][i];            }            else            t[p].r=i,t[i].val=g[p][i];            Build(i);        }    }}///*void gao(int p){    if(t[p].l==0)return ;    int l=t[p].l,r=t[p].r;    gao(l);    gao(r);    ans[p][1]=max(g[l][p],g[r][p]);    int d=g[l][p]+g[r][p];    for (int i=2;i<=m;++i)    {        ans[p][i]=max(ans[l][i-1]+g[l][p] , ans[r][i-1]+g[r][p]);        for (int j=0;j<=i-2;++j)        ans[p][i]=max(ans[p][i] , ans[l][j]+ans[r][i-j-2]+d);    }}*/int gao(int p,int num){    if(ans[p][num]!=-1)return ans[p][num];    if(!p || !num){        ans[p][num]=0;        return 0;    }    ans[p][num]=0;    for(int i=0;i<=num-1;i++){        int ls=gao(t[p].l,i);        int rs=gao(t[p].r,num-i-1);        if(ans[p][num]<ls+rs)ans[p][num]=ls+rs;    }    ans[p][num]+=t[p].val;    return ans[p][num];}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        int i,j,k;        memset(g,-1,sizeof(g));        memset(vis,0,sizeof(vis));        memset(ans,-1,sizeof(ans));        for(i=1;i<n;i++){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            g[a][b]=g[b][a]=c;        }        Build(1);        //gao(1);        //printf("%d\n",ans[1][m]);        printf("%d\n",gao(1,m));    }    return 0;}