[HDU2191] 悼念512汶川大地震遇难同胞——珍惜现在，感恩生活 [单调队列][多重背包]

[ $\frak{Link}$]

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$\mathcal{f_j=max\{f_{j-kcost_i}+kval_i\},\;k\in[0,count_i]}$
做一点微小的工作
$\mathcal{f_{r+jcost_i}=max\{f_{r+kcost_i}+(j-k)val_i\},\;k\in[j-count_i,j]}$
提成两部分
$\mathcal{f_{r+jcost_i}=max\{f_{r+kcost_i}-kval_i\}+jval_i,\;k\in[j-count_i,j]}$
由此。
枚举 $\mathcal{r}$，枚举 $\mathcal{k}$，单调队列。
当然这道题并不用优化（
*不要把题目给的东西的意义弄反
*不要忘记输出回车
*k-min(k,c)

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cctype>
using namespace std;
int T, n, m;
int p, h, c;
int f[105];
pair<int, int> q[105];
int qt, qh;
int main() {
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &m);
		for (int i = 0; i <= n; ++i) f[i] = 0;
		for (int i = 1; i <= m; ++i) {
			scanf("%d%d%d", &p, &h, &c);
			for (int j = 0; j < p; ++j) {
				qt = 0;
				qh = 1;
				for (int x, k = 0; /*qwq*/ ; ++k) {
					if ((x = k * p + j) > n) break;
					while (qh <= qt && q[qh].second < k - min(k, c)) ++qh;
					while (qh <= qt && q[qt].first < f[x] - k * h) --qt;
					q[++qt] = make_pair(f[x] - k * h, k);
					f[x] = max(f[x], q[qh].first + k * h); //放在后面防止越界 
				}
			}
		}
		printf("%d\n", f[n]);
	}
	return 0;
}