E1. Weights Division (easy version)

E1 - Weights Division (easy version)

题意

给你n个点和最高费用S，然后给你n-1条边，和每条边的边权。让你求如果可以$w_i:=\frac{w_i}{2}$，需要最少多少步可以满足$cost\le S$。

思路

先$dfs$求出每个边的边权和经过每条边的次数。

然后优先队列按{ 如果对这条边进行操作会减少多少元 }从大到小的顺序。

然后贪心即可。

每条边的下面的节点代表的是经过这条边的次数。

每条边的下面的节点代表这条边的权值。

#include 
#define INF 0x3f3f3f3f
#define DOF 0x7f7f7f7f
#define endl '\n'
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(case, x) cout << case << "  : " << x << endl
#define open freopen("ii.txt", "r", stdin)
#define close freopen("oo.txt", "w", stdout)
#define IO                       \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0)
#define pb push_back
using namespace std;
#define int long long
#define lson rt << 1
#define rson rt << 1 | 1
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> PII;
const int maxn = 1e6 + 10;
struct edge{
    int v,w;
};

vector<vector<edge>>G;
vector<int>val, cnt;

void dfs(int u, int fa = -1) {
    if(fa!=-1&&G[u].size()==1)
        cnt[u]=1;
    for(auto t : G[u]) {
        if(t.v == fa)continue;
        val[t.v]=t.w;
        dfs(t.v, u);
        cnt[u] += cnt[t.v];
    }
}

struct node{
    int cnt,val;
    bool operator<(const node t)const{
        return val*cnt-(val/2)*cnt<t.val*t.cnt-(t.val/2)*t.cnt;
    }
};
void solve() {
    int n, s;
    cin >> n >> s;
    G = vector<vector<edge>>(n + 1);
    cnt = val = vector<int>(n + 1);
    for(int i = 0; i < n-1; ++i) {
        int v, u, w;
        cin >> v >> u >> w;
        G[v].push_back({u, w});
        G[u].push_back({v, w});
    }
    dfs(1, -1);
    priority_queue<node>q;
    int sum=0;
    for(int i=2;i<=n;++i){
        sum+=cnt[i]*val[i];
        q.push({cnt[i],val[i]});
    }
    int ans=0;
    while(sum > s) {
        ++ans; node now=q.top(); q.pop();
        sum-=now.val*now.cnt-(now.val/2)*now.cnt;
        q.push({now.cnt,now.val/2});
    }
    cout<<ans<<endl;
}


signed main() {
    int t;
    cin >> t;
    while(t--) {
        solve();
    }

}