Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51939    Accepted Submission(s): 21876 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   Recommend lcy   |   We have carefully selected several similar problems for you:   1203  2159  2955  1171  2191  题目大意：经典的背包问题，先给了你n个物品的价值，再给你n个物品的体积，还给了你背包的容积，问背包选择性的装物品后最大价值是多少？ 解题思路：01背包的裸题。 代码如下： 二维实现： #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int dp[1010][1010];//dp[i][j]表示前i件物品装在容量为j的背包的总价值（前i件不一定全部装） int v[1010];//体积 int val[1010];//价值 int main() { int t; scanf("%d",&t); while(t--) { int n,V; scanf("%d%d",&n,&V); for(int i=1;i<=n;i++) { scanf("%d",&val[i]); } for(int i=1;i<=n;i++) { scanf("%d",&v[i]); } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=0;j<=V;j++) { if(v[i]<=j)//表示第i件物品能装下 { dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+val[i]);//装下的话，再判断装还是不装 } else { dp[i][j]=dp[i-1][j];//装不下，那就不装 } } } printf("%d\n",dp[n][V]); } return 0; } 滚动数组实现（因为这里只用到了前一种状态）： #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int dp[2][1010];//dp[i][j]表示前i件物品装在容量为j的背包的总价值（前i件不一定全部装） int v[1010];//体积 int val[1010];//价值 int main() { int t; scanf("%d",&t); while(t--) { int n,V; scanf("%d%d",&n,&V); for(int i=1;i<=n;i++) { scanf("%d",&val[i]); } for(int i=1;i<=n;i++) { scanf("%d",&v[i]); } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=0;j<=V;j++) { if(v[i]<=j)//表示第i件物品能装下 { dp[i%2][j]=max(dp[(i-1)%2][j],dp[(i-1)%2][j-v[i]]+val[i]);//装下的话，再判断装还是不装 } else//装不下，那就不装 { dp[i%2][j]=dp[(i-1)%2][j]; } } } printf("%d\n",dp[n%2][V]); } return 0; }一维实现： #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int dp[1010]; int v[1010]; int val[1010]; int main() { int t; scanf("%d",&t); while(t--) { int n,V; scanf("%d%d",&n,&V); for(int i=1;i<=n;i++) { scanf("%d",&val[i]); } for(int i=1;i<=n;i++) { scanf("%d",&v[i]); } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { for(int j=V;j>=v[i];j--)//倒着写，要不然会出现重复放置同一物品的情况 { dp[j]=max(dp[j],dp[j-v[i]]+val[i]); } } printf("%d\n",dp[V]); } return 0; }