先上题目：
Bone Collector
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 23533 Accepted
Submission(s): 9535
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 ?
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.
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
这一题说白了就是一条01背包的模板，只是因为物品数目最多有1000种，所以要么使用滚动数组，要么只用一维即可。滚动数组比较好实现，所以这里我用一维数组实现，但这里需要注意的是，如果是使用一位数组实现，因为原本的状态转移方程是：
max{dp[i-1][j],dp[i-1][j-c[i]]+v[i]} c[i]<=j
dp[i][j]=dp[i-1][j]
c[i]>j
0 i==0 || j==0
消去一维以后因为需要用到j-c[i]的数据，所以外面的循环是从1~n，里面的循环是从C~0(n是物品的数目，C是背包的总体积)，只有这样做j-c[i]的数据才不会在被使用之前被覆盖。
上代码：
1 #include <cstdio> 2 #include <cstring> 3 #define max(x,y) (x > y ? x : y) 4 #define MAX 1001 5 using namespace std; 6 7 int dp[MAX],n,s; 8 int v[MAX],c[MAX]; 9 10 int main() 11 { 12 int t; 13 //freopen("data.txt","r",stdin); 14 scanf("%d",&t); 15 while(t--){ 16 scanf("%d %d",&n,&s); 17 for(int i=1;i<=n;i++){ 18 scanf("%d",&v[i]); 19 } 20 for(int i=1;i<=n;i++){ 21 scanf("%d",&c[i]); 22 } 23 memset(dp,0,sizeof(dp)); 24 for(int i=1;i<=n;i++){ 25 for(int j=s;j>=0;j--){ 26 if(c[i]<=j) dp[j]=max(dp[j],(dp[j-c[i]]+v[i])); 27 //printf("%d ",dp[j]); 28 } 29 //printf("\n"); 30 } 31 printf("%d\n",dp[s]); 32 } 33 return 0; 34 }