Bone Collector——01背包

think:
1 01背包
2 注意c[]数组和w[]数组的输入顺序

sdut原题链接

Bone Collector
Time Limit: 1000MS Memory Limit: 65536KB

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).

Example Input
1
5 10
1 2 3 4 5
5 4 3 2 1

Example Output
14

Hint
hdoj2602

Author

以下为accepted代码

#include <stdio.h>#include <string.h>#define Max(a, b) (a > b? a: b)int main(){    int T, n, v, c[1004], w[1004], dp[1004];    scanf("%d", &T);    while(T--)    {        memset(dp, 0, sizeof(dp));        scanf("%d %d", &n, &v);        for(int i = 0; i < n; i++)        {            scanf("%d", &w[i]);        }        for(int i = 0; i < n; i++)        {            scanf("%d", &c[i]);        }        for(int i = 0; i < n; i++)        {            for(int j = v; j >= c[i]; j--)            {                dp[j] = Max(dp[j], dp[j-c[i]] + w[i]);            }        }        printf("%d\n", dp[v]);    }    return 0;}/***************************************************User name: Result: AcceptedTake time: 0msTake Memory: 120KBSubmit time: 2017-02-18 16:44:33****************************************************/