Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 70139 Accepted Submission(s): 29244
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 ?
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.
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
#include<iostream> #include<cstdio> #include<algorithm> #include<string.h> int val[1002][1002]; using namespace std; int main() { int T; scanf_s("%d", &T); while (T--) { int N, V; scanf_s("%d%d", &N, &V); for (int i = 0; i <= 1000; i++) for (int j = 0; j <=1000; j++) val[i][j] = 0; int w[1002], v[1002]; for (int i = 1; i <= N; i++) scanf_s("%d", &w[i]); for (int i = 1; i <= N; i++) scanf_s("%d", &v[i]); for (int i = 1; i <= N; i++) { for (int j =0; j <= V; j++) //物品所占空间可能为0，循环从零开始 { if(j>=v[i])val[i][j] = max(val[i - 1][j], val[i - 1][j- v[i]] + w[i]);
//判断是否要装v[i]; else val[i][j] = val[i - 1][j]; 装不下v[i] } } printf("%d\n", val[N][V]); } return 0; }