题目：
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 2 31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
Author
Teddy
Source
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RE: 01背包基础题：
//感觉这个比较好理解： 46ms #include#include #include #define max(a, b) a>b?a:busing namespace std;int dp[1010][1010], val[1010], vol[1010];int main(){ int t; scanf("%d", &t); while(t--) { int n, v; 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]); for(int i = 1; i <= n; i++) { for(int j = 0; j <= v; j++) //容量； { if(j < vol[i]) dp[i][j] = dp[i-1][j]; else dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]] + val[i]); } } printf("%d\n", dp[n][v]); } return 0; }
//比二维数组稍微快点； 31ms #include#include #include #define max(a, b) a>b?a:b using namespace std;int dp[1010], val[1010], vol[1010];int main(){ int t; scanf("%d", &t); while(t--) { int n, v; scanf("%d %d", &n, &v); for(int i = 1; i <= n; i++) scanf("%d", &val[i]); for(int j = 1; j <= n; j++) scanf("%d", &vol[j]); memset(dp, 0, sizeof(dp)); for(int k = 1; k <= n; k++) { for(int Q = v; Q >= vol[k]; Q--) { dp[Q] = max(dp[Q], dp[Q - vol[k]] + val[k]); } } printf("%d\n", dp[v]); } return 0; }