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BNUOJ 6727 Bone Collector

Bone Collector

1000ms
32768KB
 
This problem will be judged on HDU. Original ID: 2602
64-bit integer IO format: %I64d      Java class name: Main
 
 
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

15 101 2 3 4 55 4 3 2 1
 

Sample Output

14
 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest
 
 
解题:0-1背包。。。
 
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <ctype.h> 8 #include <cmath> 9 #include <algorithm>10 #define LL long long11 using namespace std;12 int v[1001],w[1001],dp[1001];13 int main(){14     int kase,i,j,n,m;15     scanf("%d",&kase);16     while(kase--){17         memset(dp,0,sizeof(dp));18         scanf("%d %d",&n,&m);19         for(i = 1; i <= n; i++)20             scanf("%d",v+i);21         for(i = 1; i <= n; i++)22             scanf("%d",w+i);23         for(i = 1; i <= n; i++){24             for(j = m; j >= w[i]; j--){25                 dp[j] = max(dp[j],dp[j-w[i]]+v[i]);26             }27         }28         printf("%d\n",dp[m]);29     }30     return 0;31 }
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