Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 31374 | Accepted: 13954 |
Description
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 weightWi (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.
Input
* 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
Output
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6 1 4 2 6 3 12 2 7
Sample Output
23這道題是裸的01揹包,比較簡單,核心是01揹包的狀態轉移方程,只要與動態規劃相關,下意識就覺得很難,其實,弄明白核心之後,也挺簡單的。
思路:01揹包~
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 5000
using namespace std;
int f[3*MAXN]; //狀態數組
int d[MAXN]; //每個物品的價值
int w[MAXN]; //每個物品的重量
int N,M;
int main()
{
scanf("%d%d",&N,&M);
memset(f,0,sizeof(f));
memset(d,0,sizeof(d));
memset(w,0,sizeof(w));
for(int i=1;i<=N;i++)
{
scanf("%d%d",&w[i],&d[i]);
}
for(int i=1;i<=N;i++)
{
for(int j=M;j>=w[i];j--)
{
f[j]=max(f[j],f[j-w[i]]+d[i]);//對於每個物品,取其 不放這個物品的價值或放這個物品後 中的最大值
//,如果放,所能承受的容量減去放進去的物品的重量表示放進去了這個物品
}
}
printf("%d\n",f[M]);
return 0;
}