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Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51862    Accepted Submission(s): 21839 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 一维数组：

#include<cstdio> #include<algorithm> #include<cstring> using namespace std; int N,V; int value[1010]; int volume[1010]; int dp[1010]; int main() { int t; scanf("%d",&t); while(t--) { scanf("%d %d",&N,&V); for(int i=0;i<N;i++) scanf("%d",&value[i]); for(int i=0;i<N;i++) scanf("%d",&volume[i]); memset(dp,0,sizeof(dp)); for(int i=0;i<N;i++) { for(int j=V;j>=volume[i];j--) { dp[j]=max(dp[j],dp[j-volume[i]]+value[i]); } } printf("%d\n",dp[V]); } return 0; } 二维数组： #include<cstdio> #include<algorithm> #include<cstring> using namespace std; int N,V; int val[1010]; int vol[1010]; int dp[1010][1010]; int main() { int t; scanf("%d",&t); while(t--) { 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",&vol[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]; if(j>=vol[i]) { dp[i][j]=max(dp[i-1][j],dp[i-1][j-vol[i]]+val[i]); } } /* for(int j=vol[i];j<=V;j++) 错误的写法，当 j<vol[i] 的时候没有继承上一步的值 { dp[i][j]=max(dp[i-1][j],dp[i-1][j-vol[i]]+val[i]); }*/ } printf("%d\n",dp[N][V]); } return 0; }