Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 60469 Accepted Submission(s): 25209
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
Author
Teddy
Source
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602
题意：一个叫做Bone Collector的男的有一个包，往包里放东西，使得其价值最大。
输入：注意是先输入的是价值，后是体积。
分析：01背包裸题，注意格式输入输出就行了，我就是输入格式写错了，找错误找了一个小时，QAQ
下面给出AC代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 int n; 6 int v[1010],dp[1010],d[1010];//v代表体积，d代表谷歌值 7 while(scanf("%d",&n)!=EOF) 8 { 9 while(n--) 10 { 11 memset(v,0,sizeof(v)); 12 memset(dp,0,sizeof(dp)); 13 memset(d,0,sizeof(d)); 14 int x,y; 15 cin>>x>>y; 16 for(int i=1;i<=x;i++) 17 cin>>d[i]; 18 for(int i=1;i<=x;i++) 19 cin>>v[i]; 20 for(int i=1;i<=x;i++)//01背包主函数 21 { 22 for(int j=y;j>=v[i];j--) 23 { 24 dp[j]=max(dp[j],dp[j-v[i]]+d[i]); 25 } 26 } 27 printf("%d\n",dp[y]); 28 } 29 } 30 return 0; 31 }