hdu2602Bone Collector ——动态规划(0/1背包问题)
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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 ?
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.
#include<iostream>
#include<cstring>
using namespace std;
const int v=1000+10;
const int num=1000+10;
int value[num][2]=0;
int dp[num][v]=0;
void solve(int s,int n)
for(int i=0;i<n;i++)
for(int j=0;j<=s;j++) //这道题之前一直wrong answer,找了很久都没发现问题,最后在此处找到了根源,原先写的是int j=1;后来改为0就对了,体积竟然可以是0,我也是无语了
if(j>=value[i][0])
dp[i+1][j]=max(dp[i][j],dp[i][j-value[i][0]]+value[i][1]);
else
dp[i+1][j]=dp[i][j];
int main()
int n,s,t;
cin>>t;
while(t--)
cin>>n>>s;
memset(dp,0,sizeof(dp));
memset(value,0,sizeof(value));
for(int i=0;i<n;i++)
cin>>value[i][1];
for(int i=0;i<n;i++)
cin>>value[i][0];
solve(s,n);
cout<<dp[n][s]<<endl;
return 0;
找到递推关系即可。分情况:1)若可重复使用物品,则是dp[i+1][j]=max(dp[i][j],dp[i+1][j-value[i][0]]+value[i][1])
2)若不可重复使用,则为dp[i+1][j]=max(dp[i][j],dp[i][j-value[i][0]]+value[i][1])
区别是从本行找还是从上一行找
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