Bone Collector（01背包问题入门）

Bone Collector

原题连接：点击打开链接

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15290    Accepted Submission(s): 6055

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

 

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此题的易错点在于，物品的体积可为0，于是当背包剩余空间为0时，还是可能装物品的。
AC Code：

#include <iostream>#include <string>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <map>#include <vector>#include <queue>#include <stack>#define LL long long#define MAXI 2147483647#define MAXL 9223372036854775807#define eps (1e-8)#define dg(i) cout << "*" << i << endl;using namespace std;int vol[1001], val[1001], maxVal[1001][1001];int max (const int a, const int b) {return a > b ? a : b;}int main(){    int T, N, V, ans;    scanf("%d", &T);    while(T--)    {        scanf("%d %d", &N, &V);        for(int i = 1; i <= N; ++i) scanf("%d", &val[i]);        for(int i = 1; i <= N; ++i) scanf("%d", &vol[i]);        if(V == 0)        {            ans = 0;            for(int i = 1; i <= N; ++i)               if (!vol[i]) ans += val[i];        }        else if(N == 0)  ans = 0;        else        {            for(int i = 1; i <= N; ++i)            {                for(int j = 0; j <= V; ++j)                {                    //if(i == 0 || j == 0)  maxVal[i][j] = 0;                    //else……                    /*上面的语句是错误的，当j等于0时，表明此时背包容量为0，按理说不能装任何东西，故而                    maxVal[i][0] = 0，但是，在这题中，物品的体积可为0，故此背包容量为0但是还是可以装物品。                    可以先设maxVal[i][0] = 0，但是由于使用if-else语句，设maxVal[i][0] = 0之后就不执行了，                    无法更新maxVal[i][0] = 0*/                    if(vol[i] > j)  maxVal[i][j] = maxVal[i - 1][j];                    else  maxVal[i][j] = max(maxVal[i - 1][j], maxVal[i - 1][j - vol[i]] + val[i]);                }            }            ans = maxVal[N][V];        }        printf("%d\n", ans);    }    return 0;}