HDU2602 Bone Collector

                                                
Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21408    Accepted Submission(s): 8610

Problem DescriptionMany 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 ?

 

InputThe 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. 

OutputOne integer per line representing the maximum of the total value (this number will be less than 231). 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

AuthorTeddy 

SourceHDU 1st “Vegetable-Birds Cup” Programming Open
Contest  

Recommendlcy

解題思路：本題為典型01背包問題，直接用01背包求解即可。（01背包相關內容點這裡：《背包問題九講》）

         解釋狀態轉移方程：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]);語意解析：對於第i件物品和背包剩餘容量為j時的最佳值（骨頭價值最大）=max{對於第i-1件物品和背包剩餘容量為j時的最佳值（第i件不放入背包），對於第i-1件物品和背包剩餘容量為（j-第i件物品的重量）時的最佳值+第i件物品的價值（第i件放入背包）}

        可以用滾動數組解答，原始方程可化為：dp[j]=dp[j]>(dp[j-wei[i]]+val[i])?dp[j]:(dp[j-wei[i]]+val[i]);當處理第i件物品時，dp[j]，dp[j-wei[i]]中儲存的值就是dp[i-1][j]，dp[i-1][j-wei[i]]的值，所以在處理時(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];  //骨頭價值，骨頭重量    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]);     //價值輸入        for(i=0;i<n;i++)            scanf("%d",&wei[i]);    //重量輸入        memset(dp,0,sizeof(dp));    //數組初始化        for(i=0;i<n;i++)          //動歸            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;}