hdu 2602(dp)，hdu2602dp
題目鏈接 ：http://acm.hdu.edu.cn/showproblem.php?pid=2602
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32499 Accepted Submission(s): 13379
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
題意 ：給你一個包的體積，每塊骨頭的價值和占用的體積，求出可以放入價值最大方案的價值。
分析 ：簡單的01背包，純屬模板題，也是我做的第一題背包，就直接貼代碼了。
![]()
1 #include <cstdio>
2 #include <cmath>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6
7 //dp[i][j]表示放入第i塊骨頭並占用j體積的最大價值，
8 //c[i]表示第i塊骨頭的體積
9 //w[i]表示第i塊骨頭的價值
10
11 int dp[1111][1111],c[1111],w[1111];
12 int T,N,V;
13
14 int main ()
15 {
16 int i,j;
17 scanf ("%d",&T);
18 while (T--)
19 {
20 scanf ("%d%d",&N,&V);
21 for (i=1; i<=N; i++)
22 scanf ("%d",&w[i]);
23 for (i=1; i<=N; i++)
24 scanf ("%d",&c[i]);
25 memset(dp, 0, sizeof(dp));
26 for (i=1; i<=N; i++)
27 {
28 for (j=0; j<=V; j++)
29 {
30 dp[i][j] = dp[i-1][j]; //這裡主要考慮j小於c[i]時放不下第i塊骨頭
31 if (j >= c[i])
32 dp[i][j] = max(dp[i-1][j], dp[i-1][j-c[i]]+w[i]);
33 }
34 }
35 printf ("%d\n",dp[N][V]);
36 }
37 return 0;
38 }
View Code
