0-1背包问题：
n个商品（单个商品不可拆分），放入容量为 V 的购物车，如何才能让购物车中的商品总价最大？
书中介绍的动态规划解法是设 dp[i][j] 为前 i 号商品在容量为 j 的购物车中的最大价值，并自底向上求解。这种方法比较紧凑，但是不那么自然，难以想到利用商品编号 i 和容量 j 建立动态规划数组。
下面介绍一种思路较为自然直白的方法：
设 dp[i][j] 为第 i 号~第 j 号商品放入容量 V 的购物车所能产生的最大总价。vdp[i][j] 为 dp[i][j] 对应的商品占用购物车的容量。
则有：
1、dp[i][j] = 0 ; i == j && volume_i > V --- 只有一个商品且放不下
2、dp[i][j] = value_i ; i == j && volume_i <= V --- 只有一个商品且能放下
3、dp[i][j] = 0 ; j - 1 == 1 && volume_i > V && volume_j > V
4、dp[i][j] = value_i + value_j ; j - 1 == 1 && volume_i + volume_j <= V
5、dp[i][j] =value_i ; j - 1 == 1 && volume_i <= V && volume_j > V
6、dp[i][j] =value_j ; j - 1 == 1 && volume_i > V && volume_j <= V
7、dp[i][j] = max(dp[i][k-1] + dp[k+1][j] + 0), i < k < j ; j - 1 > 1 && volume_k > V - (vdp[i][k-1] + vdp[k+1][j])
8、dp[i][j] = max(dp[i][k-1] + dp[k+1][j] + value_k), i < k < j ; j - 1 > 1 && volume_k <= V - (vdp[i][k-1] + vdp[k+1][j])
这些情况看起来很复杂，但是我们可以稍作整理：
dp[i][j] = condition(i, V); i == j
dp[i][j] = value_i + value_j ; j - 1 == 1 && volume_i + volume_j <= V
dp[i][j] = max(condition(i, V), condition(j, V)) ; j - 1 == 1 && volume_i + volume_j > V
dp[i][j] = max(dp[i][k-1] + dp[k+1][j] + condition(k, V - vdp[i][k-1] - vdp[k+1][j]))
其中
condition(k, vRemain) = 0; volume_k > vRemain
condition(k, vRemain) = value_k; volume_k <= vRemain
至此，已找到递归式，可以自底向上求解，代码如下：
测试用例，对于5个商品，放入容量为10的购物车
价值：6 3 5 4 6
体积：2 5 4 2 3
输出最大价值为17，占用体积为9
1 #include "iostream" 2 #include "vector" 3 #include "algorithm" 4 5 #define SIZE 10 6 #define INF 65535 7 8 using namespace std; 9 10 class Goods 11 { 12 public: 13 Goods(int value, int volume) 14 { 15 this->value = value; 16 this->volume = volume; 17 } 18 public: 19 int value; 20 int volume; 21 }; 22 23 pair<int, int> condition(vector<Goods> &gArr, int k, int vRemain) 24 { 25 if (gArr.at(k).volume > vRemain) 26 { 27 return make_pair(0, 0); 28 } 29 else 30 { 31 return make_pair(gArr.at(k).value, gArr.at(k).volume); 32 } 33 } 34 35 void package(vector<Goods> &gArr, int volume) 36 { 37 int dp[SIZE][SIZE], vdp[SIZE][SIZE]; 38 int size = gArr.size(); 39 pair<int, int> result1, result2; 40 41 for (int begin = 0; begin < size; begin++) 42 { 43 for (int step = 0, i = 0, j = begin; step < size - begin; step++, i++, j++) 44 { 45 if (i == j) 46 { 47 result1 = condition(gArr, i, volume); 48 dp[i][j] = result1.first; 49 vdp[i][j] = result1.second; 50 } 51 else if (abs(i - j) == 1) 52 { 53 if (gArr.at(i).volume + gArr.at(j).volume <= volume) 54 { 55 dp[i][j] = gArr.at(i).value + gArr.at(j).value; 56 vdp[i][j] = gArr.at(i).volume + gArr.at(j).volume; 57 } 58 else 59 { 60 result1 = condition(gArr, i, volume); 61 result2 = condition(gArr, j, volume); 62 dp[i][j] = result1.first > result2.first ? result1.first : result2.first; 63 vdp[i][j] = (result1.first == 0 && result2.first == 0) ? 0 : 64 (result1.first > result2.first ? result1.second : result2.second); 65 } 66 } 67 else 68 { 69 for (int k = i + 1; k < j; k++) 70 { 71 result1 = condition(gArr, k, volume - (vdp[i][k - 1] + vdp[k + 1][j])); 72 dp[i][j] = dp[i][k - 1] + dp[k + 1][j] + result1.first; 73 vdp[i][j] = vdp[i][k - 1] + vdp[k + 1][j] + result1.second; 74 } 75 } 76 } 77 } 78 79 cout << dp[0][size - 1] << endl; 80 cout << vdp[0][size - 1] << endl; 81 } 82 83 84 void main() 85 { 86 vector<Goods> gArr; 87 int maxVolume = 10; 88 89 gArr.push_back(Goods(6, 2)); 90 gArr.push_back(Goods(3, 5)); 91 gArr.push_back(Goods(5, 4)); 92 gArr.push_back(Goods(4, 2)); 93 gArr.push_back(Goods(6, 3)); 94 95 package(gArr, maxVolume); 96 }
