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 ?
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 2 31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
动态规划背包问题,滚动数组优化
#include <bits/stdc++.h> using namespace std; struct node{ int v, m; }a[1010]; int n, v; long long int dp[1010]; int main() { int t; scanf("%d", &t); while(t--){ scanf("%d%d", &n, &v); for(int i = 0; i<n; i++) scanf("%d", &a[i].m); for(int i = 0; i<n; i++) scanf("%d", &a[i].v); memset(dp, 0, sizeof(dp)); for(int i = 0; i<n; i++) for(int j = v; j >= a[i].v; j--) dp[j] = max(dp[j], dp[j - a[i].v] + a[i].m); printf("%lld ", dp[v]); } return 0; }