hdu 2602 Bone Collector
2014-08-20 00:45
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 30095 Accepted Submission(s): 12389
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 1
Sample Output
14/*题解:(对01背包问题的个人理解) 01背包问题,0代表不放,1代表放,即是对第i个物品放或者不放的背包问题。 01背包问题通常具有以下特征: 1.物品有很多种,但是每种仅有一个 2.背包容量有限,题目要求得到最优解 3.通过递推的方式,可以求得最优解 */ [code]#include<cstdio> #include<cstring> int dp[1010][1010]; int max(int a,int b) { return a>b?a:b; } int main() { int T,n,v,i,j,w[1010],val[1010]; scanf("%d",&T); while(T--) { memset(dp,0,sizeof(dp)); scanf("%d %d",&n,&v); for(i=1; i<=n; i++) scanf("%d",&val[i]); for(i=1; i<=n; i++) scanf("%d",&w[i]); for(i=1; i<=n; i++) { for(j=0; j<=v; j++) { if(j>=w[i]) { //放还是不放 dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+val[i]); } else dp[i][j]=dp[i-1][j]; } } printf("%d\n",dp [v]); } return 0; }
/*题解:
滚动数组(一维数组),解01背包问题。
参考自《算法竞赛入门经典第二版》p273
*/
#include<cstdio>
#include<cstring>
int dp[1010];
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int T,n,v,i,j,w[1010],val[1010];
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d %d",&n,&v);
for(i=1; i<=n; i++)
scanf("%d",&val[i]);
for(i=1; i<=n; i++)
scanf("%d",&w[i]);
for(i=1; i<=n; i++)
{
for(j=v; j>=0; j--)
{
if(j>=w[i]) dp[j]=max(dp[j],dp[j-w[i]]+val[i]);
}
}
printf("%d\n",dp[v]);
}
return 0;
}
[/code]
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