HDOJ-2602 Bone Collector【01背包】

题目链接HDU-2602

题目描述

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

思路

裸的01背包,没啥说的。

AC代码:

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#include <bits/stdc++.h>
using namespace std;

#define long long long
#define MAX_N 100010
#define INF 0x3f3f3f3f

int v[1010], w[1010], dp[MAX_N];


int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);

int t;
cin >> t;
while(t--)
{
memset(dp, 0, sizeof(dp));
int n, T;
cin >> n >> T;
for(int i = 1; i <= n; i++)
{
cin >> v[i];
}
for(int i = 1; i <= n; i++)
{
cin >> w[i];
}
for(int i = 1; i <= n; i++)
{
for(int j = T; j >= w[i]; j--)
{
dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
}
}
cout << dp[T] << endl;
}

return 0;
}
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