\(\mathscr{Description}\)
Link.
01 背包。
物品种类 \(n\le10^5\)，背包容量 \(m\le3\times10^5\)，单个物品体积 \(w\in\{1,2,3\}\)，价值 \(c\le10^9\)。
\(\mathscr{Solution}\)
模拟赛 T3（不是说这题）什么降智选择结构，我直接开摆。
Motivation: 只有两种体积，我会 two-pointers。
构造一发，把 \(w=1\) 的物品转化成 \(w=2\) 的物品：枚举 \(w=1\) 选择的奇偶性，若为奇，最大者必选，其余两两打包；若为偶，直接两两打包。化归到 \(w=\{2,3\}\) 的情况，算上排序的复杂度就能 \(\mathcal O(n\log n)\) 做了。
騞然已解，如土委地。（
这种拆物品、组合物品的构造 trick 还挺厉害的，要多留意一下√
\(\mathcal {Code}\)
/*+Rainybunny+*/
#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = l, rep##i = r; i <= rep##i; ++i)
#define per(i, r, l) for (int i = r, per##i = l; i >= per##i; --i)
typedef long long LL;
inline char fgc() {
static char buf[1 << 17], *p = buf, *q = buf;
return p == q && (q = buf + fread(p = buf, 1, 1 << 17, stdin), p == q) ?
EOF : *p++;
}
template <typename Tp = int>
inline Tp rint() {
Tp x = 0, s = fgc(), f = 1;
for (; s < '0' || '9' < s; s = fgc()) f = s == '-' ? -f : f;
for (; '0' <= s && s <= '9'; s = fgc()) x = x * 10 + (s ^ '0');
return x * f;
}
const int MAXN = 1e5;
int n, m;
std::vector<int> buc[3];
inline LL solve() {
std::sort(buc[1].begin(), buc[1].end(), std::greater<int>());
LL ret = 0, sum = 0;
int p = 0, q = 0;
while (p < buc[2].size() && (p + 1) * 3 <= m) sum += buc[2][p++];
while (q < buc[1].size() && (q + 1) * 2 + p * 3 <= m)
sum += buc[1][q++];
ret = std::max(ret, sum);
while (~--p) {
sum -= buc[2][p];
while (q < buc[1].size() && (q + 1) * 2 + p * 3 <= m)
sum += buc[1][q++];
ret = std::max(ret, sum);
}
return ret;
}
int main() {
n = rint(), m = rint();
rep (i, 1, n) {
int w = rint(), c = rint();
buc[w - 1].push_back(c);
}
std::sort(buc[0].begin(), buc[0].end(), std::greater<int>());
std::sort(buc[2].begin(), buc[2].end(), std::greater<int>());
auto tmp(buc[1]);
for (int i = 0; i + 1 < buc[0].size(); i += 2) {
buc[1].push_back(buc[0][i] + buc[0][i + 1]);
}
LL ans = solve();
if (buc[0].size()) {
buc[1] = tmp;
for (int i = 1; i + 1 < buc[0].size(); i += 2) {
buc[1].push_back(buc[0][i] + buc[0][i + 1]);
}
--m, ans = std::max(ans, solve() + buc[0][0]);
}
printf("%lld\n", ans);
return 0;
}