动态规划之 01背包（一维，二维）

例题 HDU 2602 Bone Collector
传送门

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
1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output
14

Author
Teddy

裸的背包问题，就是n个物品，每个都有w[i]的重量,v[i]的价值,W重量的背包可以装的最大价值是多少

最简单的思路肯定是一个递归，一种情况是拿，一种是不拿，看那个的总价值最高，但这时间复杂度是O(2^n)肯定是无法ac，因为有大量的数据被重复计算，如果我们把这些数据提前存储起来，就能把时间复杂度降到O(nw)

代码如下

#include
#include
#include
#include
#include
#include
#include

using namespace std;
const int MAX = 1010;
int w[MAX],v[MAX];
int dp[MAX][MAX];
int t,n,W;
int get(int i,int j){//表示第i个背包，空间剩余为j时的价值
    if(dp[i][j] >= 0){
        return dp[i][j];
    }
    int res;
    if(i == n)  res = 0;//当到达最后一个时，没有背包可以选择了,
    else if(j < w[i]){
        res = get(i+1,j);//不选这个背包
    }
    else res = max(get(i+1,j),get(i+1,j-w[i]) + v[i]);//返回选择或不选择这个背包中的最大的一个价值

    dp[i][j] = res;//存储数据
    return res;
}
int main(void){

    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&W);
        for(int i=0;iscanf("%d",&v[i]);
        }
        for(int i=0;iscanf("%d",&w[i]);
        }
        memset(dp,-1,sizeof(dp));
        int res = get(0,W);//表示第i个背包，当前剩余重量空间
        printf("%d\n",res);
    }

    return 0;
}

递推公式的如下

#include
#include
#include
#include
#include

using namespace std;
const int MAX = 1010;
int v[MAX],w[MAX];
int dp[MAX][MAX];
int t,n,W;
int main(void){
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&W);
        for(int i=0;iscanf("%d",&v[i]);
        for(int i=0;iscanf("%d",&w[i]);
        //dp[i][j]表示从前i个背包，j为最大重量选取的最大价值
        for(int i=0;ifor(int j=0;j<=W;j++){
                if(j < w[i])
                    dp[i+1][j] = dp[i][j];
                else
                    dp[i+1][j] = max(dp[i][j],dp[i][j-w[i]] + v[i]);//选或者不选中最大的一个
            }
        }
        printf("%d\n",dp[n][W]);
    }
    return 0;
}

背包压缩为一维数组。
核心代码如下：

void solve(){
    for(int i=1;i<=N;++i){
        for(int j=W;j>=0;--j){
            //如果j从0开始，那么如果在后面使用dp[j-w[i]]时，就不是上一次的，而是本次更新过的。
            dp[j] = max(dp[j],dp[j-w[i]] + v[i]);
        }
    }
}