Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …

The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the
maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.

Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third
line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

给定容量V的背包 每一块骨头的体积和价值固定 之后求背包能装下的最大价值和

01背包问题就是 一次性使用骨头 方法常见的有二维数组和一维数组 一维数组是二维数组再空间上的优化 循环数组优化

先解释下二维数组的方法 状态转移方程是

dp[i][v] = max(dp[i-1][v], dp[i-1][v-c[i]]+w[i])

将前i件物品放入容量为v的背包中 是子问题 若只考虑第i件物品的策略（放或不放）那么就可以转化为一个只牵扯前i-1件物品的问题

实现代码如下

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1010;
int va[maxn], vo[maxn];
int dp[maxn][maxn];
int n, v;
int res;

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &v);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &va[i]);
        }
        for(int i = 1; i <= n; ++i){
            scanf("%d", &vo[i]);
        }
        memset(dp, 0, sizeof(dp));

        for(int i = 1; i <= n; ++i){
            for(int j = 0; j <= v; ++j){
                if(vo[i] <= j)
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-vo[i]]+va[i]);
                else
                    dp[i][j]=dp[i-1][j];
            }
        }

        printf("%d\n", dp[n][v]);
    }
    return 0;
}

接下来我们说一下一维数组的方法 递推公式

for i <- 1 to N

do for v <- V to 0

do  dp[v] = max(dp[v], dp[v-c[i]]+w[i])

这里因为用到了滚动数组的感觉 所以第二重循环要逆序 这样才能保证每一次dp的更新是从上一层i-1更新得到

代码实现如下

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1010;
int va[maxn], vo[maxn];
int dp[maxn];
int n, v;
int res;

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &v);
        for(int i = 0; i < n; ++i){
            scanf("%d", &va[i]);
        }
        for(int i = 0; i < n; ++i){
            scanf("%d", &vo[i]);
        }
        memset(dp, 0, sizeof(dp));

        for(int i = 0; i < n; ++i){
            for(int j = v; j >= vo[i]; --j){
                dp[j] = max(dp[j], dp[j-vo[i]] + va[i]);
            }
        }

        printf("%d\n", dp[v]);
    }
    return 0;
}

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