HDU 2602 Bone Collector
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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
15 101 2 3 4 55 4 3 2 1
Sample Output
14
基础01背包问题
#include<stdio.h>int main(){ int v, t, n, i, j, max; int a[1005], b[1005]; scanf("%d", &t); while (t--) { scanf("%d%d", &n, &v); for (i = 1; i <= n; i++) scanf("%d", &a[i]); for (i = 1; i <= n; i++) scanf("%d", &b[i]); int f[1005] = { 0 }; for (i = 1; i <= n;i++) for (j = v; j >= b[i]; j--) f[j] = f[j] > f[j - b[i]] + a[i] ? f[j] : f[j - b[i]] + a[i]; for (max = 0, i = 0; i <= v; i++)max = max < f[i] ? f[i] : max; printf("%d\n", max); } return 0;}
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