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HDU 2602 Bone Collector(01背包)
Bone Collector
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
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#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 1005; int n, v, d[N], val[N], vol[N], cas; int main() { scanf ("%d", &cas); while (cas--) { memset (d, 0, sizeof (d)); 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 j = v; j >= vol[i]; --j) d[j] = max (d[j], d[j - vol[i]] + val[i]); } printf ("%d\n", d[v]); } return 0; }
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