[SCOI2007]组队

嘟嘟嘟

这题有人说部分分O(n3)暴力，然而我暴力都没写过，调了半天也没用……还是看题解吧

首先，咱把A * ( h – minH ) + B * ( s – minS ) <= C 变个型，得到 A * h + B * s - C <= A * minH + B * minS. 令 sum = A * h + B * s - C,如果我们把所有球员按sum排序，就能保证取球员的时候是单调的，如果 i 能取，则 j (j < i) 一定能取。

然后我们第一层循环枚举minS,第二层循环枚举minH,然后设两个指针L, R,表示当前符合sum <= A * minH + B * minS的区间，但同时我们还要保证h >= minH && s >= minS,而且根据h >= minH还要保证，s <= minS + C / B.于是就有一个很【强】的做法，我们先把符合的条件的s添加进去，再把队列中不符合条件的h踢出。

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<cctype>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 5e3 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) last = ch, ch = getchar();
     while(isdigit(ch)) ans = (ans << ) + (ans << ) + ch - '', ch = getchar();
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) putchar('-'), x = -x;
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n;
 ll A, B, C;
 struct Node
 {
     int h, s; ll sum;
 }Sum[maxn], H[maxn], S[maxn];
 bool cmp1(Node a, Node b) {return a.sum < b.sum;}
 bool cmp2(Node a, Node b) {return a.h < b.h;}
 bool cmp3(Node a, Node b) {return a.s < b.s;}

 int ans = ;

 int main()
 {
     n = read(); A = read(); B = read(); C = read();
     for(int i = ; i <= n; ++i)
     {
         Sum[i].h = read(), Sum[i].s = read(); Sum[i].sum = (ll)A * Sum[i].h + (ll)B * Sum[i].s - C;
         H[i].h = Sum[i].h; H[i].s = Sum[i].s; H[i].sum = Sum[i].sum;
         S[i].h = Sum[i].h; S[i].s = Sum[i].s; S[i].sum = Sum[i].sum;
     }
     sort(Sum + , Sum + n + , cmp1);
     sort(H + , H + n + , cmp2);
     sort(S + , S + n + , cmp3);
     for(int i = ; i <= n; ++i)        //枚举minS
     {
         int L = , R = , cnt = ;
         ll Mins = S[i].s;
         ll Lims = Mins + C / B;
         for(int j = ; j <= n; ++j)        //枚举minH
         {
             ll Limsum = A * H[j].h + B * Mins;
             while(R < n && Sum[R + ].sum <= Limsum)    //合法区间，但不能保证s，h符合
             {
                 R++;
                 if(Mins <= Sum[R].s && Sum[R].s <= Lims) cnt++;    //符合条件的s
             }
             while(L < n && H[L + ].h < H[j].h)        //维护合法区间左端点
             {
                 L++;
                 if(Mins <= H[L].s && H[L].s <= Lims) cnt--;    //踢出不符合的h
                 //因为[L, R]中可能有cnt之外的，所以要判断哪些属于cnt,踢出时再cnt--
             }
             ans = max(ans, cnt);
         }
     }
     write(ans); enter;
     return ;
 }