POJ 3624 Charm Bracelet
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 31350 | Accepted: 13943 |
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 theN (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weightWi (1 ≤ Wi ≤ 400), a 'desirability' factorDi (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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
Source
简单O1背包水题，温习一下，瞬秒。
此题智能用一维DP数组，二维妥妥超内存。
一维AC
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN=3405;
const int MAXM=12885;
int dp[MAXM];
int weight[MAXN];
int value[MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)>0)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%d%d",&weight[i],&value[i]);
for(int i=1;i<=n;i++)
for(int j=m;j>=weight[i];j--)
dp[j]=max(dp[j-weight[i]]+value[i],dp[j]);
printf("%d\n",dp[m]);
}
return 0;
}
二维数组，方法正确，但此题会超内存
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int MAXN=3405;
const int MAXM=12885;
int dp[MAXN][MAXM];
int weight[MAXN];
int value[MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)>0)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
scanf("%d%d",&weight[i],&value[i]);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
dp[i][j]=dp[i-1][j];
if(j>=weight[i])
dp[i][j]=max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]);
}
printf("%d\n",dp[n][m]);
}
return 0;
}