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 2 31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
>>题目链接<<
题意:给定背包容量,骨头的个数和每个骨头的价值,求在背包容量允许的情况下,最多装的价值。
思路:最基础的01背包,dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
#include<cstdio> #include<iostream> using namespace std; int main() { int N,V,T; int w[1005]; int c[1005]; int i,j,k; scanf("%d",&T); while(T--) { int f[1005]={0}; scanf("%d%d",&N,&V); 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++) { // printf("\nc[%d]=%d\tw[%d]=%d\n",i,c[i],i,w[i]); for(int v=V;v>=c[i];v--)// 剩下的体积要大于想放入的体积 { f[v]=max(f[v],f[v-c[i]]+w[i]); // printf("v=%d\tf[%d]=%d\n",v,v,f[v]); } } } return 0; } /* 1 5 10 2 3 4 5 1 4 3 2 1 5*/