背包问题-极大极小值-空间恰好是j

一、01背包

1、最小值

求价值最小值：初始化(f[0][0] = 0), 其余是(INF)

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品只能选一个，求总体积恰好是(m)的最小价值

输入
4 5
1 2
2 4
3 4
4 5
输出
7
二维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    cin >> n >> m;
    memset(f, INF, sizeof f);
    f[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = min(f[i][j], f[i - 1][j - v] + w);
        }
    }
    cout << f[n][m] << endl;
    return 0;
}

一维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    cin >> n >> m;
    memset(f, INF, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = m; j >= v; j--)
            f[j] = min(f[j], f[j - v] + w);
    }
    cout << f[m] << endl;
    return 0;
}

2、最大值

求价值最大值：初始化(f[0][0] = 0), 其余是(-INF)

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品只能选一个，求总体积恰好是(j)的最大价值

输入
4 5
1 2
2 4
3 4
4 5
输出
8
二维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    cin >> n >> m;

    memset(f, -INF, sizeof f);
    f[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = max(f[i][j], f[i - 1][j - v] + w);
        }
    }
    cout << f[n][m] << endl;
    return 0;
}

一维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    cin >> n >> m;

    memset(f, -INF, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = m; j >= v; j--) {
            f[j] = max(f[j], f[j - v] + w);
        }
    }
    cout << f[m] << endl;
    return 0;
}

二、完全背包

1、最小值

求价值最小值：初始化(f[0][0] = 0), 其余是(INF)

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品可以选无数多个，求总体积恰好是(m)的最小价值

输入
4 5
1 2
2 4
3 4
4 5
输出
7
二维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    cin >> n >> m;

    memset(f, INF, sizeof f);
    f[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = min(f[i][j], f[i][j - v] + w);
        }
    }
    cout << f[n][m] << endl;
    return 0;
}

一维

#include <bits/stdc++.h>

using namespace std;
const int N = 110, INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    cin >> n >> m;

    memset(f, INF, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = v; j <= m; j++) {
            f[j] = min(f[j], f[j - v] + w);
        }
    }
    cout << f[m] << endl;
    return 0;
}

2、最大值

求价值最大值：初始化(f[0][0] = 0), 其余是(-INF)

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品可以选无数多个，求总体积恰好是(m)的最大价值

输入
4 5
1 2
2 4
3 4
4 5
输出
10

二维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    cin >> n >> m;

    memset(f, -INF, sizeof f);
    f[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = max(f[i][j], f[i][j - v] + w);
        }
    }
    cout << f[n][m] << endl;
    return 0;
}

一维

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    cin >> n >> m;

    memset(f, -INF, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        cin >> v >> w;
        for (int j = v; j <= m; j++) {
            f[j] = max(f[j], f[j - v] + w);
        }
    }
    cout << f[m] << endl;
    return 0;
}