【HDU2602】Bone Collector(01背包入门级)

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 2 31).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

思路：01背包就是对每个物体进行放或者不放的操作。
状态转移方程： dp[j]=max(dp[j],dp[j-weight]+price[i]);

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 1e4+5;struct FYJ{    int v;    int w;} fyj[maxn];int dp[maxn];int max( int a,int b ){    return a>b? a:b;}int main(){    int T;    int n,m;    scanf("%d",&T);    while( T-- )    {        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&m);        for( int i=1 ; i<=n ; i++ ) scanf("%d",&fyj[i].v);        for( int i=1 ; i<=n ; i++ ) scanf("%d",&fyj[i].w);        for( int i=1 ; i<=n ; i++ )           for( int j=m ; j>=fyj[i].w ; j-- )           {            dp[j] = max( dp[j] , dp[j-fyj[i].w]+fyj[i].v );           }           printf("%d\n",dp[m]);    }    return 0;}