F. Gardening Friends

原题链接

题解

最大深度·k-与节点1的距离·c
其中最大深度只要知道了节点1的最大深度，其子节点的最大深度可分类讨论得出

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

vector<ll> G[200005];
ll height1[200005] = {0};
ll height2[200005] = {0};
ll rec[200005] = {0};
ll cnt[200005] = {0};

inline void read(ll &x) {
	x = 0;
	ll flag = 1;
	char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-')flag = -1;
        c = getchar();
    }
	while(c >= '0' && c <= '9') {
		x = (x << 3) + (x << 1) + (c ^ 48);
		c = getchar();
	}
	x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
    	putchar('-');
		x = -x;
	}
    if(x > 9)
		write(x / 10);
    putchar(x % 10 + '0');
}

void dfs1(ll now, ll fa)
{
    height1[now] = height2[now] = 0;
    for(auto next : G[now])
    {
        if(next == fa) continue;

        dfs1(next, now);
        if(height1[next] + 1 > height1[now])
        {
            height1[now] = height1[next] + 1;
        }
        else if(height1[next] + 1 > height2[now])
        {
            height2[now] = height1[next] + 1;
        }
    }
}

void dfs2(ll now, ll fa, ll op)
{
    rec[now] = height1[now];
    cnt[now] = op;
    for(auto next : G[now])
    {
        if(next == fa) continue;

        ll tem2 = height1[next];
        if(height1[next] + 1 == height1[now])
        {
            ll tem1 = height1[now];
            height1[next] = max(height1[next], height2[now] + 1);
            height1[now] = height2[now];
            dfs2(next, now, op + 1);
            height1[now] = tem1;
            height1[next] = tem2;
        }
        else
        {
            height1[next] = height1[now] + 1;
            dfs2(next, now, op + 1);
            height1[next] = tem2;
        }
    }
}

int main()
{
    ll t;
    read(t);
    while(t--)
    {
        ll n, k, c;
        read(n); read(k); read(c);
        for(ll i = 1; i < n; i++)
        {
            ll x, y;
            read(x); read(y);
            G[x].push_back(y);
            G[y].push_back(x);
        }

        dfs1(1, 1);
        dfs2(1, 1, 0);
        ll ans = 0;
        for(ll i = 1; i <= n; i++)
        {
            ans = max(ans, rec[i] * k - cnt[i] * c);
            G[i].clear();
        }
        write(ans);
        putchar('\n');
    }
    return 0;
}