HDOJ-2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11188 Accepted Submission(s): 4317


[align=left]Problem Description[/align]
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 ?



[align=left]Input[/align]
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.

[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).

/*
//0-1背包问题，容量可以从前向后推（二维数组），也可以从后向前推（一维数组）

//代码一：
#include<stdio.h>
#include<string.h>
int main()
{
int t,V,i,j,N;
int val[1001],vol[1001],record[1001];
scanf("%d",&t);
while(t--)
{
memset(record,0,sizeof(record));
scanf("%d%d",&N,&V);
for(i=0;i<N;++i)
scanf("%d",&val[i]);
for(i=0;i<N;++i)
scanf("%d",&vol[i]);
for(i=0;i<N;++i)
for(j=V;j>=vol[i];--j)
if(record[j-vol[i]]+val[i]>record[j])
record[j]=record[j-vol[i]]+val[i];
printf("%d\n",record[V]);
}
return 0;
}

//代码二：
#include<stdio.h>
#include<string.h>
int main()
{
int t,V,i,j,N;
int val[1001],vol[1001],record[1001];
scanf("%d",&t);
while(t--)
{
memset(record,0,sizeof(record));
scanf("%d%d",&N,&V);
for(i=0;i<N;++i)
scanf("%d",&val[i]);
for(i=0;i<N;++i)
{
scanf("%d",&vol[i]);
for(j=V;j>=vol[i];--j)
if(record[j-vol[i]]+val[i]>record[j])
record[j]=record[j-vol[i]]+val[i];
}
printf("%d\n",record[V]);
}
return 0;
}
*/

//代码三：   二位数组的写法
#include<stdio.h>
#include<string.h>

int val[1001],vol[1001];
int dp[1001][1001];
int main()
{
int T,i,j,n,v;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&v);
for(i=1;i<=n;++i)
scanf("%d",&val[i]);
for(i=1;i<=n;++i)
scanf("%d",&vol[i]);
memset(dp,0,sizeof(dp));
for(i=1;i<=n;++i)
for(j=0;j<=v;++j)     //注意这里要从0开始
if(j>=vol[i])     //背包容量能过装下第i件骨头的情况下选取装与不装的最大值
dp[i][j]=dp[i-1][j]>dp[i-1][j-vol[i]]+val[i]?dp[i-1][j]:dp[i-1][j-vol[i]]+val[i];
else               //背包容量不能装下第i件的化当前最大值为装下前i-1件骨头的状态
dp[i][j]=dp[i-1][j];
printf("%d\n",dp
[v]);
}
return 0;
}