题目描述
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
有N件物品和一个容量为V的背包。第i件物品的重量是c[i],价值是w[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量,且价值总和最大。
输入输出格式
输入格式:
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
第一行:物品个数N和背包大小M
第二行至第N+1行:第i个物品的重量C[i]和价值W[i]
输出格式:
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
输出一行最大价值。
输入输出样例
4 6 1 4 2 6 3 12 2 7
23
题解:背包qwq不须解释
#include<cstdio> #include<iostream> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> typedef long long ll; using namespace std; #define maxn 3403 #define maxm 12881 int m,n,w[maxn],c[maxn],f[maxm]; int main() { scanf("%d%d",&n,&m); for (int i=1;i<=n;++i) scanf("%d%d",&w[i],&c[i]); for (int i=1;i<=n;++i) for (int v=m;v>=w[i];--v) if (f[v-w[i]] + c[i] > f[v]) f[v] = f[v-w[i]]+c[i]; printf("%d",f[m]); return 0; }