0-1背包问题

经典的0-1背包问题。
代码中的backtracking部分是输出组成最大价值的编号。

// runtime: 4ms
// space: 728K
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
int main() {

    int c, n; // c for count, n for number;
    while (cin>> c >> n) {
        int dp[n + 1][c + 1];
        int price[n + 1], mark[n + 1];
        for (int i = 1; i <= n; ++i) {
            cin >> price[i] >> mark[i];
        }

        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= c; ++j) {
                if (i == 0 || j == 0) {
                    dp[i][j] = 0;
                    continue;
                }
                if (j < price[i]) {
                    dp[i][j] = dp[i - 1][j];
                    continue;
                }
                else {
                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - price[i]] + mark[i]);
                }
            }
        }

        // output dp
        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= c; ++j) {
                cout << dp[i][j] << " ";
                if (j == c)
                    cout << endl;
            }
        }

        cout << dp[n][c] << endl;

        // Backtracking
        int row = n, col = c;
        stack<int> s;
        while (row > 0 && col > 0) {
            if (dp[row][col] == dp[row - 1][col]) { // not selected
                row--;
                continue;
            } else { // selected
                s.push(row);
                col -= price[row];
                row--;
            }
        }

        while (!s.empty()) {
            cout << s.top() << endl;
            s.pop();
        }
    }
    return 0;
}

空间优化为一维数组。

// runtime: 4mm
// space; 528K
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main() {
    int c, n; // c for capacity, n for number
    while (scanf("%d %d", &c, &n) != EOF) {
        int V[n], P[n]; // V for value, P for Price
        for (int i = 0; i < n; ++i) {
            scanf("%d %d", &P[i], &V[i]);
        }

        int dp[c + 1];
        memset(dp, 0, sizeof(dp)); // initialize dp.

        for (int i = 0; i < n; ++i) {
            for (int j = c; j >= P[i]; --j) {
                dp[j] = max(dp[j], dp[j - P[i]] + V[i]);
            }
        }
        printf("%d\n", dp[c]);
    }
    return 0;
}