Bread

Bread

题解：哈夫曼树模型

该题目是给出根节点，求出其WPL,但是我们要仔细阅读，题目说面包是可以被剩余的，所以我们只要将剩余的面包当成子节点放进队列即可

#include <bits/stdc++.h>
#define Zeoy std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0)
#define all(x) (x).begin(), (x).end()
#define endl '\n'
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const int N = 2e5 + 10;
ll qpow(ll x, ll y)
{
    ll ans = 1;
    while (y)
    {
        if (y & 1)
            ans *= x;
        x *= x;
        y >>= 1;
    }
    return ans;
}
priority_queue<ll, vector<ll>, greater<ll>> q;
int main(void)
{
    Zeoy;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        ll n, l;
        cin >> n >> l;
        for (int i = 1; i <= n; ++i)
        {
            ll x;
            cin >> x;
            q.push(x);
            l -= x;
        }
        if (l > 0)				//将多余的面包当成子节点放入队列
            q.push(l);
        ll sum = 0L;
        while (q.size() != 1)
        {
            ll t = 0;
            t += q.top();
            q.pop();
            t += q.top();
            q.pop();
            q.push(t);
            sum += t;
        }
        cout << sum << endl;
    }
    return 0;
}