杭電 2602 Bone Collector（背包問題 ）
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29109 Accepted Submission(s): 11898
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 2
31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
Author
Teddy
Source
HDU 1st “Vegetable-Birds Cup”
Programming Open Contest
這是 01 背包問題 01 背包問題用到了遞推的思想，並且用到了分治
將大問題轉化為能夠容易解決的，小問題，從小問題來得到答案的思想
大致思路如下：
c[i-1][v]表示前i-1件物品恰放入一個質量為m的背包可以取到最大價值，若只考慮第i件物品的話，那麼有兩種情況，就是放，與不放，要是放的話，那麼問題就轉化為“將前n-1件物品放入剩下的容量為m-w[i]的背包中，此時獲得的最大價值就是c[i-1][m-w[i]]再加上最後放進去的第i件的物品所獲得價值p[i];要是不放的話，問題就變成：”將前n-1件物品放入容量為j的背包中，此時獲得的最大價值就是c[i-1][m];然後題上一般有物品的數量，及相應的價值，通過兩個for循環就好了；最後的最大價值就是c[n][m];
代碼如下：
#include
#include
int dp[1010][1010],val[1010],vol[1010];
int main()
{
int n,i,j,v,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&v);
for(i=1;i<=n;i++)
scanf("%d",&val[i]);
for(j=1;j<=n;j++)
scanf("%d",&vol[j]);
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
for(j=0;j<=v;j++)//題上有個陷阱，就是將沒有體積但是有價值的 骨頭考慮了進去
{
if(vol[i]<=j&&dp[i-1][j]
