Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39259    Accepted Submission(s): 16261

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 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602 题意：给出n件物品的重量和价值，放进一个容量为v的背包，使背包里的价值最大。 0-1背包的模板题，可以有两种解法。 一维数组： #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int w[1100],p[1110]; int f[1110]; int main() {int t,n,v;scanf("%d",&t); while(t--){scanf("%d%d",&n,&v);for(int i=1;i<=n;i++)scanf("%d",&w[i]); //输入物品重量 for(int i=1;i<=n;i++)scanf("%d",&p[i]); //输入物品价值 memset(f,0,sizeof(f));for(int i=1;i<=n;i++) {for(int j=v;j>=w[i];j--) //这个循环保证了放进去的物品重量不会超过背包所能容纳的重量 {if(f[j] < f[j-w[i]] + p[i]) // 如果当前所拥有价值 小于 加上这件物品时创造的价值就更新 f[j]= f[j-w[i]] + p[i]; // f[j] 表示背包重量为 j 时背包里的最大价值，// 所以f[ j - w[i] ] 表示放进这件物品时的状态（因为放进该件物品后容量就减少了） }} printf("%d\n",f[v]);} return 0; }

二维数组：

1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 int w[1100],p[1110]; 6 int f[1110][1110]; 7 int main() 8 { 9 int t,n,v; 10 scanf("%d",&t); 11 while(t--) 12 { 13 scanf("%d%d",&n,&v); 14 for(int i=1;i<=n;i++) 15 scanf("%d",&p[i]); 16 for(int i=1;i<=n;i++) 17 scanf("%d",&w[i]); 18 memset(f,0,sizeof(f)); 19 for(int i=1;i<=n;i++) 20 { 21 for(int j=0;j<=v;j++) 22 { 23 if(w[i]<=j) // 这件物品的重量小于当前的容量，也就是说放的进背包 24 { 25 // f[i][j] 表示第 i 件物品在背包容量为 j 时的状态， 26 //所以 f[i-1][j] 表示背包在上一次容量为 j 时候的状态，也就是没放这件物品的时候 27 f[i][j]=max(f[i-1][j],f[i-1][j-w[i]]+p[i]);// 比较 没放进去之前 和放进该物品后 的价值，取最大 28 29 } 30 else f[i][j]=f[i-1][j]; // 如果不能放进该物品，则取上一次的状态 31 } 32 } 33 printf("%d\n",f[n][v]); 34 } 35 return 0; 36 }

 

渣渣一枚，如果有什么不对的地方，还请各位大神批评指正~  (^_^)

转载于:https://www.cnblogs.com/ember/p/4701035.html

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