一眼题,由于每条路径只删去一个点是最优情况,用一个树形dp枚举这个删这个点的答案和不删这个点的答案即可

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#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<cstdlib>
#include<ctime>

typedef long long ll;

using namespace std;

namespace mstd{

inline void qread(int &x)
{
x=0; int f=1;
static char c=getchar();
while(!isdigit(c)) {if(c=='-') f=-f; c=getchar();}
while(isdigit(c)) x=(x*10+c-'0'),c=getchar();
x*=f;
}

inline void qread(long long &x)
{
x=0; long long f=1;
static char c=getchar();
while(!isdigit(c)) {if(c=='-') f=-f; c=getchar();}
while(isdigit(c)) x=(x*10+c-'0'),c=getchar();
x*=f;
}
}

const int maxn=100100;

int n,head[maxn],size,f[maxn],rt;

struct edge{
int next,to,dis;
}e[maxn];

inline void addedge(int next,int to,int dis)
{
e[++size].to=to;
e[size].dis=dis;
e[size].next=head[next];
head[next]=size;
}

void dfs(int t,int fat)
{
int i,j,k;
int tmp=0;
for(i=head[t];i;i=e[i].next)
{
j=e[i].to;
k=e[i].dis;
if(j==fat) continue;
f[j]=k;
dfs(j,t);
tmp+=f[j];
}
if(tmp) f[t]=min(tmp,f[t]);
}

int main()
{
int i,j;
memset(f,0x3f,sizeof(f));
mstd::qread(n);
mstd::qread(rt);
int t1,t2,t3;
for(i=1;i<n;i++)
mstd::qread(t1),mstd::qread(t2),mstd::qread(t3),addedge(t1,t2,t3),addedge(t2,t1,t3);
dfs(rt,rt);
printf("%d\n",f[rt]);
return 0;
}