【洛谷 2871】手链

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

输入样例#1：  复制 4 6 1 4 2 6 3 12 2 7 输出样例#1：  复制 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; }

 

转载于:https://www.cnblogs.com/wuhu-JJJ/p/11182336.html