例题:一个商店里有5件商品,重量分别为:2、3、4、5、9,对应的价值分别为:3、4、5、8、10。现在,有一个小偷拿着一个20kg的包去这个商店里偷东西,问:他如何偷,才能使他不虚此行,偷的价值总和最大?(这年头,小偷不会动态规划都不行了,哈哈哈)
其中k代表前k个商品,C代表剩余多少空间
c++代码实现:
#include<iostream>
#define N 6
#define W 21
using namespace std;
int B[N][W] = {0};
int w[6] = {0,2,3,4,5,9};
int v[6] = {0,3,4,5,8,10};
void package()
{
int k, C;
for(k=1;k<N;k++) {
for (C = 1; C < W; C++) {
if (w[k] > C) {
B[k][C] = B[k - 1][C];
} else {
int value1 = B[k - 1][C - w[k]] + v[k];
int value2 = B[k - 1][C];
if (value1 > value2) {
B[k][C] = value1;
} else {
B[k][C] = value2;
}
}
}
}
}
int main()
{
package();
cout<<B[5][20]<<endl;
return 0;
}
python3 代码实现:
weight = [0, 2, 3, 4, 5, 9]
price = [0, 3, 4, 5, 8, 10]
N = 6
W = 21
B = [[0 for i in range(W)] for j in range(N)]
for k in range(1, N):
for left in range(1, W):
if weight[k] > left:
B[k][left] = B[k - 1][left]
else:
value1 = B[k - 1][left - weight[k]] + price[k]
value2 = B[k - 1][left]
if value1 > value2:
B[k][left] = value1
else:
B[k][left] = value2
print(B[5][20])
整理自:https://www.bilibili.com/video/av36136952/?spm_id_from=333.788.videocard.3