Bone Collector



A - Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

 15 101 2 3 4 55 4 3 2 1 

 

Sample Output

 14 

 

这个题就属于典型的01背包问题。

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int dp[10000];int bonev[10000];         //记录每个物品的价值int bonew[10000];         //记录每个物品的体积int max(int a,int b){    return a>b?a:b;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,v,i,j;        memset(dp,0,sizeof(dp));        scanf("%d %d",&n,&v);        for(i=1; i<=n; i++)            scanf("%d",&bonev[i]);        for(i=1; i<=n; i++)            scanf("%d",&bonew[i]);        for(i=1; i<=n; i++)       //遍历每个物品            for(j=v; j>=bonew[i]; j--)       //对于01背包问题，体积是从大到小遍历，而完全背包则是从小到大遍历                 dp[j]=max(dp[j-bonew[i]]+bonev[i],dp[j]);        printf("%d\n",dp[v]);    }    return 0;}