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]);

}

View Code

 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;

}

View Code

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;

}

View Code

 dp[i][j]表示重量恰好为j的最大价值