#2478. 「九省聯考 2018」林克卡特樹
阿新 • • 發佈:2019-02-08
這題挺考思路的…我這種渣渣就是做不來.
大佬blog
想了半天,然後看題解了半天…思路還是看大佬的吧.
c++程式碼如下:
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i = x ; i <= y; ++ i)
#define repd(i,x,y) for(register int i = x ; i >= y; -- i)
using namespace std;
typedef long long ll;
template<typename T>inline void read(T&x)
{
x = 0;char c;int sign = 1;
do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c));
do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c));
x *= sign;
}
const ll inf = 1e12;const int N = 3e5+50;
int n,k;ll K;
int head[N],to[N << 1],nxt[N << 1],w[N << 1 ],tot;
inline void add(int x,int y,int val)
{
w[tot] = val;
to[tot] = y;
nxt[tot] = head[x];
head[x] = tot++;
}
#define PI pair<ll,int>
#define mk make_pair
PI f[N][3],t[3];
PI operator + (PI a,PI b) { return mk(a.first + b.first,a.second + b.second); }
PI max(PI a,PI b)
{
if (a.first > b.first) return a;
else if(a.first == b.first && a.second < b.second) return a;
return b;
}
inline void chkmax(PI&a,PI b,int w,int op) { b.first += w - K * op;b.second += op; a = max(a,b); }
inline void update(int x,int y,int w)
{
t[0] = f[x][0]; t[1] = f[x][1]; t[2] = f[x][2];
chkmax(f[x][2],t[1] + f[y][1],w,-1);
chkmax(f[x][2],t[1] + f[y][0],w,0);
chkmax(f[x][2],t[2] + f[y][2],0,0);
chkmax(f[x][1],t[0] + f[y][0],w,1);
chkmax(f[x][1],t[0] + f[y][1],w,0);
chkmax(f[x][1],t[1] + f[y][2],0,0);
chkmax(f[x][0],t[0] + f[y][2],0,0);
}
void dfs(int x,int fa)
{
f[x][0] = mk(0,0);
f[x][1] = mk(-inf,0);
f[x][2] = mk(-inf,0);
for(register int i = head[x];~i;i = nxt[i])
if(to[i] != fa){
dfs(to[i],x);
update(x,to[i],w[i]);
}
f[x][2] = max(f[x][2],f[x][1]);
f[x][2] = max(f[x][2],f[x][0]);
f[x][2] = max(f[x][2],mk(f[x][0].first-K,f[x][0].second+1));
}
inline bool check(ll x)
{
K = x;
dfs(1,1);
return f[1][2].second <= k;
}
int main()
{
memset(head,-1,sizeof head);
read(n); read(k);++k;
rep(i,2,n)
{
int u,v,w;
read(u); read(v); read(w);
add(u,v,w); add(v,u,w);
}
ll l = -inf,r = inf,mid,ans=0;
while(l <= r)
{
if(check(mid = l + r>> 1)) ans = f[1][2].first + k * K,r = mid - 1;
else l = mid + 1;
}
cout << ans << endl;
return 0;
}