有N道题目,题目有简单与困难之分,简单的题目花费A分钟,困难的题目花费B分钟,那么考试时间一共有T的情况下,我们是可以提前交卷的,但是有些时间限制,就是譬如说你现在第x分钟交卷,但是你这时候就是必须要完成所有的t[i]≤x的题目才行,不然就使算作0分。
所以,这里直接按照限制时间t[i]升序排列来进行贪心即可。每次看看能不能补充额外的,譬如说现在的时间还能多出来做几道简单题之类的。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define _ABS(x, y) ( x > y ? (x - y) : (y - x) )
#define lowbit(x) ( x&( -x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define efs 1e-7
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 2e5 + 7;
int N, T, A, B, easy, diff, op[maxN];
struct node
{
int t, r, id;
friend bool operator < (node e1, node e2) { return e1.r == e2.r ? e1.t < e2.t : e1.r < e2.r; }
}a[maxN];
int main()
{
int Cas; scanf("%d", &Cas);
while(Cas--)
{
scanf("%d%d%d%d", &N, &T, &A, &B);
easy = diff = 0;
for(int i=1; i<=N; i++)
{
a[i].id = i;
scanf("%d", &a[i].t);
op[i] = a[i].t;
if(a[i].t) { a[i].t = B; diff++; }
else { a[i].t = A; easy++; }
}
for(int i=1; i<=N; i++) scanf("%d", &a[i].r);
int ans = 0, tmp;
ll det;
ll tim = 0, res;
sort(a + 1, a + N + 1);
a[N + 1].r = T + 1;
for(int i=1; i<=N + 1 && tim <= T; i++)
{
if(tim < a[i].r)
{
res = a[i].r - tim - 1;
tmp = i - 1;
det = min(1LL * easy, res / A);
res -= det * A;
tmp += det;
det = min(1LL * diff, res / B);
tmp += det;
ans = max(ans, tmp);
}
tim += a[i].t;
if(op[a[i].id]) diff--;
else easy--;
}
printf("%d\n", ans);
}
return 0;
}