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;
}
/*
*/