0/1背包模板题目_hdu2602

那个背包装骨头的题目，原题不重要
主要是一个应该注意的bug：物品的重量可能是0
本来应该早就学会的题目，唉。学了就不晚

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <vector>

using namespace std;

const int maxweight = 1005;
const int maxnum = 1005;
int dp[maxnum][maxweight];
struct food {
    int weight, money;
    bool operator<(food b) {
        return weight < b.weight;
    }
} f[maxnum];
int n, m;
void solve() {
    int i, j;
    for (i = 1; i <= n; i++) {
        for (j = 0; j <= m; j++) { //可能有体积0的物品，不能从1开始循环
            if (j >= f[i].weight) {
                dp[i][j] = max(dp[i - 1][j - f[i].weight] + f[i].money, dp[i - 1][j]);
            } else {
                dp[i][j] = dp[i - 1][j];
            }
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("zz_in.txt", "r", stdin);
    freopen("zz_op.txt", "w", stdout);
#endif
    int t, i, j, k;
    cin >> t;
    while (t--) {
        memset(dp, 0, sizeof(dp));
        cin >> n >> m;
        for (i = 1; i <= n; i++) {
            cin >> f[i].money;
        }
        for (i = 1; i <= n; i++) {
            cin >> f[i].weight;
        }
        solve();
        cout << dp[n][m] << endl;
        //dp数组已经存的最优解了，没必要再找一次
    }

#ifdef LOCAL
    printf("Time used = %.2f\n", (double) clock() / CLOCKS_PER_SEC);
#endif
    return 0;
}

豪华升级成滚动数组（一维数组）
需要注意：倒着循环

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <vector>

using namespace std;

const int maxweight = 1005;
const int maxnum = 1005;
// int dp[maxnum][maxweight];
int dp[maxweight];
struct food {
    int weight, money;
    bool operator<(food b) {
        return weight < b.weight;
    }
} f[maxnum];
int n, m;
void solve() {
    int i, j;
    // for (i = 1; i <= n; i++) {
    //     for (j = 0; j <= m; j++) { //可能有体积0的物品，不能从1开始循环
    //         if (j >= f[i].weight) {
    //             dp[i][j] = max(dp[i - 1][j - f[i].weight] + f[i].money, dp[i - 1][j]);
    //         } else {
    //             dp[i][j] = dp[i - 1][j];
    //         }
    //     }
    // }
    for (i = 1; i <= n; i++) {
        for (j = m; j >= 0; j--) { //倒着循环，防止需要用到的上一个数组的数据被覆盖
            if (j >= f[i].weight) {
                dp[j] = max(dp[j - f[i].weight] + f[i].money, dp[j]);
            }
        }
    }
}

int main() {
#ifdef LOCAL
    freopen("zz_in.txt", "r", stdin);
    freopen("zz_op.txt", "w", stdout);
#endif
    int t, i, j, k;
    cin >> t;
    while (t--) {
        memset(dp, 0, sizeof(dp));
        cin >> n >> m;
        for (i = 1; i <= n; i++) {
            cin >> f[i].money;
        }
        for (i = 1; i <= n; i++) {
            cin >> f[i].weight;
        }
        solve();
        // cout << dp[n][m] << endl;
        // //dp数组已经存的最优解了，没必要再找一次
        cout << dp[m] << endl;
    }

#ifdef LOCAL
    printf("Time used = %.2f\n", (double) clock() / CLOCKS_PER_SEC);
#endif
    return 0;
}