Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 77450 Accepted Submission(s): 32095

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

分析：

经典的01背包问题，每个物品只有两种状态，放还是不放

dp[i][j]:即前面i件物品放入一个容量为j的背包可以获得的最大价值

状态转移方程：

dp[i][j]=dp[i-1][j] j<w[i]

dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i] j>=w[i]

注意：

1.先输入的是物品的价值，而不是重量，题目中好像说反了

2.背包容量可以为0，物品的重量也可以为0

关于第2点的测试数据：

输入

5 0
2 4 1 5 1
0 0 1 0 0

输出

从前往后遍历二维数组

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,c;
        scanf("%d %d",&n,&c);
        int v[n+1],w[n+1];
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        int dp[n+1][c+1];
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=c;j++)
            {
                if(w[i]<=j)//表示第i个物品放入背包中
                {
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);//第i个物品放入之后，那么前面i-1个物品可能会因为剩余空间不够无法放入
                }else//表示第i个物品不放入背包
                {
                    dp[i][j]=dp[i-1][j];//如果第i个物品不放入背包，那么此时的最大价值与放前面i-1个物品的值相等
                }
            }
        }
        printf("%d
",dp[n][c]);
    }
    return 0;
}

/*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0


15
14
12
*/

从后往前遍历二维数组

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,c;
        scanf("%d %d",&n,&c);
        int v[n+1],w[n+1];
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        int dp[n+1][c+1];
        memset(dp,0,sizeof(dp));
        for(int j=0;j<=c;j++)
        {
            if(j>=w[n])
            {
                dp[n][j]=v[n];
            }else
            {
                dp[n][j]=0;
            }
        }
        for(int i=n-1;i>=1;i--)
        {
            for(int j=0;j<=c;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
",dp[1][c]);
    }
    return 0;
}

/*
3
5 10
6 3 5 4 6
2 2 6 5 4
5 10
1 2 3 4 5
5 4 3 2 1
5 0
2 4 1 5 1
0 0 1 0 0


15
14
12
*/

二者其实是一个方法，只是遍历二维数组的方向有点不同