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~ (^_^)