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 ?
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 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14就是01背包真接上代碼
#include<stdio.h> int c[3500],w[3500],f[13000]; int main() { int i,j,k,n,v,m,t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=0;i<=m;i++) f[i]=0; for(i=0;i<n;i++) scanf("%d",w+i); for(i=0;i<n;i++) scanf("%d",c+i); for (i = 0; i < n; ++i) { for (v = m; v >= c[i]; --v) //注意這裡的V是從大到小c[i]可優化為bound,bound = max {V - sum c[i,...n],c[i]} { f[v] = (f[v] > f[v - c[i]] + w[i]?f[v] : f[v - c[i]] + w[i]); } } printf("%d\n",f[m]); } return 0; } #include<stdio.h> int c[3500],w[3500],f[13000]; int main() { int i,j,k,n,v,m,t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=0;i<=m;i++) f[i]=0; for(i=0;i<n;i++) scanf("%d",w+i); for(i=0;i<n;i++) scanf("%d",c+i); for (i = 0; i < n; ++i) { for (v = m; v >= c[i]; --v) //注意這裡的V是從大到小c[i]可優化為bound,bound = max {V - sum c[i,...n],c[i]} { f[v] = (f[v] > f[v - c[i]] + w[i]?f[v] : f[v - c[i]] + w[i]); } } printf("%d\n",f[m]); } return 0; }