01背包

题目

2. 01背包问题 - AcWing题库

有N件物品和一个容量是V的背包。每件物品只能使用一次。

第i件物品的体积是vi，价值是wi。

求解: 将哪些物品装入背包，可使这些物品的总体积不超过背包容量，且总价值最大。
输出最大价值。

思路

https://alchemist-al.com/algorithms/knapsack-problem

状态转移方程

f(i, v) 表示前 i 件物品恰放入一个容量为 v 的背包可以获得的最大价值。

python3解法

方法一：二维数组

N, V = map(int, input().split())

goods = []
for i in range (N):
    goods.append([int(i) for i in input().split()])
    
dp = [[0 for i in range (V+1) ] for j in range (N+1)]
for i in range (1, N+1):
    row = dp[i]
    [v, w] = goods[i - 1]
    for j in range (1, V+1):
        if j >= v:
            row[j] = max(dp[i-1][j], dp[i-1][j-v] + w)
        else:
            row[j] = dp[i-1][j]

print(dp[N][V])

方法二：一维数组

N, V = map(int, input().split())

goods = []
for i in range (N):
    goods.append([int(i) for i in input().split()])
    
dp = [0 for i in range (V+1) ]
for i in range (1, N+1):
    [v, w] = goods[i - 1]
    for j in range (V, 0, -1):
        if j >= v:
            dp[j] = max(dp[j], dp[j-v] + w)

print(dp[V])

golang解法

方法一：二维数组

package main

import (
    "fmt"    
    "math"
)

func main() {
    var N, V int
    fmt.Scanf("%d %d", &N, &V)
    
    v := make([]int, N)
    w := make([]int, N)
    for i:=0; i<N; i++ {
        fmt.Scanf("%d %d", &v[i], &w[i])
    }

    dp := make([][]int, N+1)
    for i:=0; i<=N; i++ {
        dp[i] = make([]int, V+1)
    }
    
    for i:=1; i<=N; i++ {
        vi := v[i-1]
        wi := w[i-1]
        for j:=1; j<=V; j++ {
            if j >= vi {
                dp[i][j] = int(math.Max(float64(dp[i-1][j]), float64(dp[i-1][j-vi] + wi)))   
            } else {
                dp[i][j] = dp[i-1][j]
            }
        }
    }
    
    fmt.Println(dp[N][V])
}

方法二：一维数组

package main

import (
    "fmt"    
    "math"
)

func main() {
    var N, V int
    fmt.Scanf("%d %d", &N, &V)
    
    v := make([]int, N)
    w := make([]int, N)
    for i:=0; i<N; i++ {
        fmt.Scanf("%d %d", &v[i], &w[i])
    }

    dp := make([]int, V+1)
    
    for i:=1; i<=N; i++ {
        vi := v[i-1]
        wi := w[i-1]
        for j:=V; j>=1; j-- {
            if j >= vi {
                dp[j] = int(math.Max(float64(dp[j]), float64(dp[j-vi] + wi)))   
            }
        }
    }
    
    fmt.Println(dp[V])
}

c语言解法

方法一：二维数组

#include <stdio.h>

#define LENGTH 1001

int main() {
    int N, V;
    scanf("%d %d", &N, &V);
 
    int v[LENGTH], w[LENGTH];
    for (int i=0; i<N; i++) {
        scanf("%d %d", &v[i], &w[i]);
    }
    
    int dp[LENGTH][LENGTH];
    for (int i=1; i<=N; i++) {
        int vi = v[i-1];
        int wi = w[i-1];
        for (int j=1; j<=V; j++) {
            if (j >= vi && dp[i-1][j] < dp[i-1][j-vi] + wi) {
                dp[i][j] = dp[i-1][j-vi] + wi;
            } else {
                dp[i][j] = dp[i-1][j];
            }
        }
    }
    
    printf("%d", dp[N][V]);
    return 0;
}

方法二：一维数组

#include <stdio.h>

#define LENGTH 1001

int main() {
    int N, V;
    scanf("%d %d", &N, &V);
 
    int v[LENGTH], w[LENGTH];
    for (int i=0; i<N; i++) {
        scanf("%d %d", &v[i], &w[i]);
    }
    
    int dp[LENGTH];
    for (int i=0; i<=V; i++) {
        dp[i] = 0;
    }
    
    for (int i=1; i<=N; i++) {
        int vi = v[i-1];
        int wi = w[i-1];
        for (int j=V; j>=1; j--) {
            if (j >= vi && dp[j] < dp[j-vi] + wi) {
                dp[j] = dp[j-vi] + wi;
            }
        }
    }
    
    printf("%d", dp[V]);
    return 0;
}