HDOJ 題目2602 Bone Collector(動態規劃,01揹包)
阿新 • • 發佈:2019-02-18
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29252 Accepted Submission(s): 11970
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
Author Teddy
Source 思路:純01揹包,模板就夠 ac程式碼
#include<stdio.h> #include<string.h> int maxx(int a,int b) { if(a>b) return a; else return b; } int main() { int v[100000],w[100000],i,j,n,m,t; int dp[2000]; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) scanf("%d",&v[i]); for(i=0;i<n;i++) scanf("%d",&w[i]); for(i=0;i<n;i++) for(j=m;j>=w[i];j--) { dp[j]=maxx(dp[j],dp[j-w[i]]+v[i]); } printf("%d\n",dp[m]); } }