多校3 hdu6060 RXD and dividing

先记录每个节点与哪些节点相连接，把1当做根节点dfs一遍找到每个节点有多少子节点（包括子节点的子节点），并记录当前节点与其父节点连边的权值。

枚举2—n节点，计算其于父节点连边的贡献，相加得到答案。#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#define mod 1000000007
using namespace std;
int e;
inline long long mini(long long a, long long b)
{
if (a < b)
{
return a;
}
return b;
}
int vis[1000000+5];
struct node
{
long long son,fw;
}q[2000005];
int head[2000000+5];
struct nn
{
int v,next;
long long w;
}ed[2000000+5];
void add(int u,int v,int w)
{
ed[e].v = v; ed[e].w = w;
ed[e].next = head[u];
head[u] = e ++;
}
long long dfs(int x)
{
long long ans = 0;
int flag = 0;
vis[x] = 1;
for(int i = head[x]; i != -1; i = ed[i].next)
{
int tx = ed[i].v;
if(vis[tx]) continue;
q[tx].fw =ed[i].w;
flag =1;
ans += dfs(tx);
}
if(!flag)
{
q[x].son = 0;
return 1;
}
else
{
q[x].son = ans;
return ans+1;
}
}

int main()
{
int n;
long long m;
while(~scanf("%d%lld",&n,&m))
{
memset(q,0,sizeof(q));
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
e = 0;
for(int i = 0; i < n-1; i++)
{
int u,b,ww;
scanf("%d%d%d",&u,&b,&ww);
add(u,b,ww); add(b,u,ww);
}
dfs(1);
long long ans = 0;
for(int i = 2; i <= n; i++)
{

if(q[i].son)
ans = ans + q[i].fw*mini(m,q[i].son + 1);
else
ans += q[i].fw;
// cout << i << " " << ans <<endl;
}
printf("%lld\n",ans);
}
}