Charm Bracelet

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 weight W_i (1 ≤ W_i ≤ 400), a ‘desirability‘ factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

大意：在一共M大的背包里面放N件东西，计算这N件东西的最大价值，很基础的背包问题，即在第二个for循环中从背包的最大容量开始，因为背包可以有剩余，然后减少，用递归，dp[j] = max(dp[j],dp[j-w[i]]+v[i]);

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int maxn = 222222;
int w[maxn],v[maxn],dp[maxn];
int main(){
    int n,m;
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n ; i++)
        scanf("%d%d",&w[i],&v[i]);
        for(int i = 1; i <= n ; i++){
                for(int j = m; j >= w[i];j--){
                        dp[j] = max(dp[j],dp[j-w[i]]+v[i]);
                }
        }
        printf("%d",dp[m]);
}

二维实现，不过会超时

1

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[4000][15000];
int w[4000],v[4000];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1; i <= n; i++)
            scanf("%d%d",&w[i],&v[i]);
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n ; i++){
            for(int j = 0; j <= m ;j++){
                if(j >= w[i])
                    dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
                else dp[i][j] = dp[i-1][j];
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}

dp[i][j]代表的是重量小于等于j的最大价值

2.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[4000][15000];
int w[4000],v[4000];
const int inf = 0x3f3f3f3f;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1; i <= n; i++)
            scanf("%d%d",&w[i],&v[i]);
        for(int i = 0; i <= n ; i++){
            for(int j = 0 ; j <= m ;j++){
                dp[i][j] = -inf;
            }
        }
        dp[0][0] = 0;
        for(int i = 1; i <= n ; i++){
            for(int j = 0; j <= m ;j++){
                if(j >= w[i])
                    dp[i][j] = max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
                else dp[i][j] = dp[i-1][j];
            }
        }
        int max1 = -inf ;
        for(int i = 0; i <= m; i++)
            max1 = max(max1,dp[n][i]);
        printf("%d\n",dp[n][m]);
    }
    return 0;
}

dp[i][j]表示重量恰好为j的最大价值