洛谷__P7095 [yLOI2020] 不离（二分）

题目链接：P7095 [yLOI2020] 不离 - 洛谷

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题目大意：

有 n 件装备，穿戴第 i 件需要当前力量 ≥ a_i，精神 ≥ b_i，穿戴后力量增加 c_i，精神增加 d_i。
求：穿完所有装备所需的最小初始力量值，以及此前提下最小的初始精神值。

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思路：

先用二分确定最小的力量值，

再用小根堆和二分确定最小的精神值

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代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#include<array>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/numeric>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define mst(a,x) memset(a,x,sizeof (a))
#define gmap __gnu_pbds::gp_hash_table
#define power __gnu_cxx::power
using namespace std;
typedef pair<int, int> pii;

const int N = 200086, mod = 998244353;
int n, m;
int res1, res2;

struct node {
    int a, b, c, d;
    bool operator<(const node &it)const {
        return a < it.a;
    }
} e[N];

struct n0de {
    int b, c, d;
    bool operator<(const n0de &it)const {
        return b > it.b;
    }
};

bool check(int mid) {
    int t = mid;
    for (int i = 1; i <= n; i++) {
        if (e[i].a <= t) t += e[i].c;
        else return false;
    }
    return true;
}

bool check0(int mid) {

    priority_queue<n0de> q;
    int a1 = res1, b1 = mid;
    
    int i = 1;
    while (i <= n && e[i].a <= a1) {
        q.push({e[i].b, e[i].c, e[i].d});
        i++;
    }
    
    while (q.size()) {
        auto [b, c, d] = q.top();
        q.pop();
        
        if (b > b1) return false;     
        a1 += c;
        b1 += d;
        
        while (i <= n && e[i].a <= a1) {
            q.push({e[i].b, e[i].c, e[i].d});
            i++;
        }
        
    }
    
    return true;
}

void solve() {
    int tt;
    cin >> tt >> n;
    if (!n) {
        cout << "0 0" << endl;
        return;
    }
    
    for (int i = 1; i <= n; i++) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        e[i] = {a, b, c, d};
    }
    
    sort(e + 1, e + n + 1);
    
    int l = e[1].a - 1, r = e[n].a + 1;
    while (l + 1 < r) {
        int mid = l + r >> 1;
        if (check(mid)) res1 = r = mid;
        else l = mid;
    }
    
    l = -1, r = 1e9;
    while (l + 1 < r) {
        int mid = l + r >> 1;
        if (check0(mid)) res2 = r = mid;
        else l = mid;
    }
    
    cout << res1 << " " << res2 << endl;
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
    // cin >> T;
    while (T--) solve();
    
    return 0;
}