Problem Description
Today is CRB‘s birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency unit).
At the shop, there are N kinds of presents.
It costs Wi Won to buy one present of i -th kind. (So it costs k × Wi Won to buy k of them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies if she buys x (x> 0) presents of i -th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤ 20
1 ≤ M ≤ 2000
1 ≤ N ≤ 1000
0 ≤ Ai, Bi ≤ 2000
1 ≤ Wi ≤ 2000
Input
There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case:
The first line contains two integers M and N .
Then N lines follow, i -th line contains three space separated integers Wi , Ai and Bi .
Output
For each test case, output the maximum candies she can gain.
Sample Input
1
100 2
10 2 1
20 1 1
Sample Output
21
Hint
CRB‘s mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
题目大意像一个背包问题,物品可以放无穷次,但是增益是Ai×次数+Bi。
关键是多了B,不然就是一个普通的背包,能处理掉B,这题就能做。
于是考虑了这样一个状态p[flag][j]:
flag为真表示放过第i个物品的最优,flag为假表示不放第i个物品的最优。
j就是背包容量。
于是p[1][j] = max(p[0][j-w[i]]+a[i]+b[i], p[1][j-w[i]]+a[i]);
然后到i+1时,初始状态是都没有放过i+1物品的,所以需要初始化p[0][j]:
p[0][i] = max(p[0][i], p[1][i]);
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <algorithm> #define LL long long using namespace std; const int maxN = 1005; int n, m, p[2][2005]; int w[maxN], a[maxN], b[maxN]; void input() { scanf("%d%d", &m, &n); for (int i = 0; i < n; ++i) scanf("%d%d%d", &w[i], &a[i], &b[i]); memset(p, 0, sizeof(p)); } void work() { for (int i = 0; i < n; ++i) { for (int j = w[i]; j <= m; ++j) p[1][j] = max(p[0][j-w[i]]+a[i]+b[i], p[1][j-w[i]]+a[i]); for (int j = 0; j <= m; ++j) p[0][j] = max(p[0][j], p[1][j]); } int ans = 0; for (int i = 0; i <= m; ++i) { ans = max(ans, p[0][i]); ans = max(ans, p[1][i]); } printf("%d\n", ans); } int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 0; times < T; ++times) { input(); work(); } return 0; }