洛谷P4165 [SCOI2007]组队(排序 堆)

题意

题目链接

sol

跟我一起大喊：n方过百万，暴力踩标算！

一个很显然的思路是枚举\(h, s\)的最小值算，复杂度\(o(n^3)\)

我们可以把式子整理一下，变成

\[a h_i + b s_i \leqslant c + aminh + bmins\]

首先按\(h\)排序

考虑去从大到小枚举\(aminh\)，同时用个vector \(n^2\)维护\(s\)序列(直接\(lowerbound + insert\))

再从大到小枚举\(bmins\)，同时用堆维护\(ah_i + b_i\)，当堆顶不满足条件的时候直接弹掉即可，用堆内元素更新答案

没错在bzoj上被卡了

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#include<ext/pb_ds/priority_queue.hpp>
#define pair pair<int, int> 
#define mp make_pair
#define fi first 
#define se second 
using namespace std;
const int maxn = 5001, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a, b, c, th[maxn], ts[maxn];
struct node {
    int h, s, id;
    bool operator < (const node &rhs) const {
        return h < rhs.h;
    }
}a[maxn];
signed main() {
    n = read(); a = read(); b = read(); c = read();
    for(int i = 1; i <= n; i++) th[i] = a[i].h = read(), a[i].s = read();
    sort(th + 1, th + n + 1);
    sort(a + 1, a + n + 1); 
    int ans = 0;
    vector<pair> v;
    for(int i = n; i >= 1; i--) {
        pair now = mp(b * a[i].s, a * a[i].h + b * a[i].s);
        v.insert(lower_bound(v.begin(), v.end(), now), now);
        //__gnu_pbds::priority_queue<int> q; 
        priority_queue<int> q;
        int r = v.size() - 1; 
        for(int j = r; j >= 0; j--) {
            q.push(v[j].se);
            while(!q.empty() && q.top() > c + a * th[i] + v[j].fi) q.pop();
            ans = max(ans, (int)q.size());
        }
    }
    cout << ans;
    return 0;
}
/*
*/