Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21408 Accepted Submission(s): 8610
Total Submission(s): 21408 Accepted Submission(s): 8610
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
解题思路：本题为典型01背包问题，直接用01背包求解即可。（01背包相关内容点这里：《背包问题九讲》）
解释状态转移方程：dp[j]=dp[j]>(dp[j-wei[i]]+val[i])?dp[j]:(dp[j-wei[i]]+val[i]);
原始方程：dp[i][j]=dp[i-1][j]>(dp[i-1][j-wei[i]]+val[i])?dp[i-1][j]:(dp[i-1][j-wei[i]]+val[i]);语意解析：对于第i件物品和背包剩余容量为j时的最佳值（骨头价值最大）=max{对于第i-1件物品和背包剩余容量为j时的最佳值（第i件不放入背包），对于第i-1件物品和背包剩余容量为（j-第i件物品的重量）时的最佳值+第i件物品的价值（第i件放入背包）}
可以用滚动数组解答，原始方程可化为：dp[j]=dp[j]>(dp[j-wei[i]]+val[i])?dp[j]:(dp[j-wei[i]]+val[i]);当处理第i件物品时，dp[j]，dp[j-wei[i]]中存储的值就是dp[i-1][j]，dp[i-1][j-wei[i]]的值，所以在处理时(i=0;i<n;i++)&&(j=v;j>=0;j--)如是即可。
#include<stdio.h> #include<string.h> int main() { int val[1002],wei[1002]; //骨头价值，骨头重量 int dp[1002]; int t,n,v; int i,j; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&v); for(i=0;i<n;i++) scanf("%d",&val[i]); //价值输入 for(i=0;i<n;i++) scanf("%d",&wei[i]); //重量输入 memset(dp,0,sizeof(dp)); //数组初始化 for(i=0;i<n;i++) //动归 for(j=v;j>=0;j--) if(j-wei[i]>=0) dp[j]=dp[j]>(dp[j-wei[i]]+val[i])?dp[j]:(dp[j-wei[i]]+val[i]); printf("%d\n",dp[v]); } return 0; }
