Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21463 Accepted Submission(s): 8633
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 ?
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.
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
Recommend
lcy
背包问题.....一定要多练习...
代码:----
1 #include<stdio.h> 2 #include<string.h> 3 #define maxn 1005 4 int dp[maxn],arr[maxn][2]; 5 int max(int a,int b) 6 { 7 return a>b?a:b; 8 } 9 10 void zeroonepack(int cost ,int value,int v) 11 { 12 for(int i=v;i>=cost;i--) 13 dp[i]=max(dp[i],dp[i-cost]+value); 14 15 } 16 int main() 17 { 18 int t,n,v ,i; 19 scanf("%d",&t); 20 while(t--) 21 { 22 scanf("%d%d",&n,&v); 23 memset(dp,0,sizeof dp); 24 for(i=0;i<n;i++) 25 { 26 scanf("%d",arr[i]+0); 27 } 28 for(i=0;i<n;i++) 29 { 30 scanf("%d",arr[i]+1); 31 } 32 for(i=0;i<n;i++) 33 zeroonepack(arr[i][1],arr[i][0],v); 34 printf("%d ",dp[v]); 35 } 36 return 0; 37 }
优化后代码:
#include<stdio.h> #include<stdlib.h> #include<string.h> struct st { int a; int b; }; typedef struct st sta; int main() { int test,n,v,i,j; scanf("%d",&test); while(test--) { scanf("%d%d",&n,&v); int *dp =(int *)malloc(sizeof(int)*(v+1)); sta *stu =(sta *)malloc(sizeof(sta)*(n+1)); for(i=0;i<n;i++) scanf("%d",&stu[i].a); for(i=0;i<n;i++) scanf("%d",&stu[i].b); for(i=0;i<=v;i++) dp[i]=0; for(i=0;i<n;i++) { for(j=v ; j>=stu[i].b ; j--) { if(dp[j]<dp[j-stu[i].b]+stu[i].a) dp[j]=dp[j-stu[i].b]+stu[i].a; } } printf("%d ",dp[v]); free(dp); free(stu); } return 0; }