01背包问题（动态规划）(DP)

https://www.bilibili.com/video/BV1jt411m7Rc 讲得很好！看这个

https://www.bilibili.com/video/BV1X741127ZM  当做补充。其实不用看。。

以"集合"角度 来分析DP问题：有限集合中，最优值。 

01背包问题

输入样例

4 5
1 2
2 4
3 4
4 5

输出样例：

8

分析&代码

https://www.bilibili.com/video/BV1jt411m7Rc 讲得很好！看这个

----------------

 

#include 
#include 
#include 
#include 
using namespace std;

int capacity;
int num;

int w[1001] = {0};  // weight
int v[1001] = {0}; // value
int f[1001][1001] = {0}; // 结果

void show() {
    for (int i = 0; i <= num ;i ++){
        printf("%d: weight:%d value:%d \n",i,w[i],v[i]);
    }
    for (int i = 0; i <= num; i++){
        for (int j = 0; j <= capacity ; j++){
            cout << f[i][j] << " ";
        }
        cout << endl;
    }

}
int main() {


    cin >> num >> capacity;
    for (int i = 1; i <= num ; i++) {
        cin >> w[i] >> v[i];
    }

    for (int i = 1; i <= num; i++) {
        for (int j = 1; j <= capacity ; j++) {
            if (w[i] > j) {
               // printf("第%d个物品 装不下，当前能装的capacity：%d \n",i,j);
                f[i][j] = f[i - 1][j];
            } else {
                int value1 = f[i - 1][j - w[i]] + v[i];
                int value2 = f[i - 1][j];

                //printf("对于第%d个物品，若装下，%d, 若不装， %d，当前能装的capacity：%d \n",i,value1,value2,j);
                f[i][j] = max(value1, value2);
            }
        }

    }
//    show();
    cout << f[num][capacity] << endl;

}