Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 28903 Accepted Submission(s): 11789
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 ?
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.
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
Author
Teddy
Source
Recommend
有N件物品和一个容量为V的背包，第i件物品的体积为v[i]，价值为w[i]，求解将哪些物品装入背包才干使总价值最大。
这是最基础的背包问题，每种物品仅仅有一件。且仅仅有两种状态：放与不放。
用子问题定义状态：dp[i][v]表示前i件物品恰好放入一个容量为m的包中可获得的最大价值。
状态转移方程：dp[i][m] = max(dp[i-1][m], dp[i-1][m-v[i]]+w[i]);
可转化为一维情况：dp[m] = max(dp[m], dp[m-v[i]]+w[i]);
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n, v, dp[1010], value[1010], volume[1010];
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &v);
for(int i = 1; i <= n; i++)
scanf("%d", &value[i]);
for(int i = 1; i <= n; i++)
scanf("%d", &volume[i]);
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; i++)
for(int j = v; j >= volume[i]; j--)
dp[j] = max(dp[j], dp[j-volume[i]]+value[i]);
printf("%d
", dp[v]);
}
return 0;
}