HDU2602 Bone Collector 【01背包】

Bone Collector

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

Total Submission(s): 28365    Accepted Submission(s): 11562

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

01背包入门题。

#include <stdio.h>#include <string.h>#define maxn 1002int dp[maxn], w[maxn], v[maxn];int main(){    int t, n, val, i, j;    scanf("%d", &t);    while(t--){        scanf("%d%d", &n, &val);        for(i = 1; i <= n; ++i) scanf("%d", v + i);        for(i = 1; i <= n; ++i) scanf("%d", w + i);        memset(dp, 0, sizeof(dp));        for(i = 1; i <= n; ++i){            for(j = val; j >= w[i]; --j){                if(dp[j] < dp[j-w[i]] + v[i])                     dp[j] = dp[j-w[i]] + v[i];            }        }        printf("%d\n", dp[val]);    }    return 0;}