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 = 1000 + 10; 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 = 1; i <= n; i++) scanf("%d", &wi[i]); for(int i = 1; i <= n; i++) scanf("%d", &vi[i]); memset(dp, 0, sizeof(dp)); //attention for(int i = 1; i <= n; i++) ZeroOnePack(vi[i], wi[i]); printf("%d ", dp[v]); } return 0; }