Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51478 Accepted Submission(s): 21677
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
- 题意:问背包最多能放下多少价值的骨头,每个骨头的价值和体积已知;
- 思路:状态转移方程:f[i][v]=max(f[i-1][v],f[i-1][v-c[i]]+w[i]);
- 失误:数据输反了;
- 代码如下:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1e3+11;
int dp[MAXN],c[MAXN],w[MAXN];
int main()
{
int N,T,V,i,j;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&N,&V);
for(i=1;i<=N;++i) scanf("%d",&w[i]);//输入反了
for(i=1;i<=N;++i) scanf("%d",&c[i]);
memset(dp,0,sizeof(dp));
for(i=1;i<=N;++i)
{
for(j=V;j>=c[i];--j)
{
dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
}
}
printf("%d\n",dp[V]);//写成N了
}
return 0;
}