题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602
题目分析：0-1背包 注意dp数组的清空， 二维转化为一维后的公式变化
/*Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 34192 Accepted Submission(s): 14066 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 Author
Teddy Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
*/
//ZeroOnePack
#include <cstdio>
#include <cstring>
const int maxn = + ;
int dp[maxn], n, v, wi[maxn], vi[maxn];
int Max(int a, int b)
{
return a > b ? a : b;
}
void ZeroOnePack(int C, int W)
{
for(int i = v; i >= C; i--)
dp[i] = Max(dp[i], dp[i-C]+W);
} int main()
{
int T;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &v);
for(int i = ; i <= n; i++) scanf("%d", &wi[i]);
for(int i = ; i <= n; i++) scanf("%d", &vi[i]);
memset(dp, , sizeof(dp)); //attention
for(int i = ; i <= n; i++)
ZeroOnePack(vi[i], wi[i]);
printf("%d\n", dp[v]);
}
return ;
}