hdu2602 Bone Collector--01背包

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=2602
一：原题内容

[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).
 
[align=left]Sample Input[/align]

1
5 10
1 2 3 4 5
5 4 3 2 1

[align=left]Sample Output[/align]

14

二：分析理解

dp
[v]表示在容积为v的情况下，放进前n个骨头的最大重量。

三：AC代码

#include<iostream>
#include<algorithm>

using namespace std;

int T;
int n, v;
int va[1005];
int vo[1005];
int dp[1005][1005];

int main()
{
scanf("%d", &T);
while (T--)
{
scanf("%d%d", &n, &v);

memset(dp, 0, sizeof(dp));

for (int i = 1; i <= n; i++)
scanf("%d", &va[i]);
for (int i = 1; i <= n; i++)
scanf("%d", &vo[i]);

for (int i = 1; i <= n; i++)
{
for (int j = 0; j <= v; j++)
{
if (j - vo[i] >= 0)
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - vo[i]] + va[i]);
else
dp[i][j] = dp[i - 1][j];
}
}

printf("%d\n", dp
[v]);
}

return 0;
}

参考博客： http://www.cnblogs.com/Su-Blog/archive/2012/08/28/2659872.html