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 ?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 1Sample Output
14**题意:容量为V的背包,尽可能装价值更多的石头,输出背包最大的价值
思路:01背包的经典套路,由石头1推到石头n,第i个石头下背包容量V的最大价值由i-1递推而来#include#include #define N 1000int num[N+10],w[N+10],v[N+10];int main(){ int i, j; int t, n, V; scanf("%d",&t); while( t --) { memset(num,0,sizeof(num)); memset(w,0,sizeof(w)); memset(v,0,sizeof(v)); scanf("%d%d",&n,&V); for(i = 1; i <= n; i ++) scanf("%d",&w[i]); for( i = 1; i <= n; i ++) scanf("%d",&v[i]); for( i = 1; i <= n; i ++) for( j = V; j >=v[i];j --)/*j代表背包的剩余容量*/ { num[j] = max(num[j],num[j-v[i]]+w[i]); /*num[j]表示不把第i个石头放入背包的价值*/ /*num[j-v[i]]+w[i]表示把第i个石头放入背包后的价值*/ /*比较二者的最优解存入num[j]*/ } printf("%d\n",num[V]); } return 0;}