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Bone CollectorTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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 2
31).
Sample Input
Sample Output
翻译:有一人爱收集骨头,每个骨头有一个价值和重量,他有一个背包,容量为V,问他装得最大价值量是多少?(经典的01背包问题)
重点:01背包、动态规划
难点:动态方程的书写
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int T,n,v,i,j,dp[1100],a[1100],b[1100];//a[1100]是价值,b[1100]是重量; cin>>T; while(T--) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(dp,0,sizeof(dp)); cin>>n>>v;//n是骨头个数,V是背包容量 for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=n;i++) scanf("%d",&b[i]); for(i=1;i<=n;i++) for(j=v;j>=b[i];j--) { if(dp[j]<dp[j-b[i]]+a[i]) dp[j]=dp[j-b[i]]+a[i]; } printf("%d\n",dp[v]); } return 0; } |