Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51885    Accepted Submission(s): 21849 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  

f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

代码： <pre name="code" class="html">#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int V[1010]; int value[1010]; int dp[1010][1010]; 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",&value[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++) { dp[i][j]=dp[i-1][j];//每次都要计算一下不放第i个物品时的价值； if(j>=V[i])//若能放第i个物品； { dp[i][j]=max(dp[i][j],dp[i-1][j-V[i]]+value[i]); } } } printf("%d\n",dp[n][v]); } return 0; }