Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 51860    Accepted Submission(s): 21837 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   0-1背包是最基础的背包问题，特点是：每种物品仅有一件，可以选择放或不放。 用子问题定义状态：即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]} 这个方程非常重要，基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下： “将前i件物品放入容量为v的背包中”这个子问题，若只考虑第i件物品的策略（放或不放），那么就可以转化为一个只牵扯前i-1件物品的问题。 在前i件物品放进容量v的背包时，它有两种情况： f[i-1][v]：如果不放第i件物品，那么问题就转化为“前i-1件物品放入容量为v的背包中”，价值为f[i-1][v]； f[i-1][v-c[i]]+w[i]：如果放第i件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-c[i]的背包中”，此时能获得的最大价值就是f[i-1][v-c[i]]再加上通过放入第i件物品获得的价值w[i]。 最后比较第一种与第二种所得价值的大小，哪种相对大，f[i][v]的值就是哪种（这里是重点，理解！）。   #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int w[1010]={0}; int c[1010]={0}; int val[1010]; int main() { int t,i,j; int N,V; scanf("%d",&t); while(t--) { memset(val,0,sizeof(val)); scanf("%d%d",&N,&V); for(i=1;i<=N;i++){ scanf("%d",&w[i]); } for(i=1;i<=N;i++){ scanf("%d",&c[i]); } for(i=1;i<=N;i++) { for(j=V;j>=c[i];j--) { val[j]=max(val[j],val[j-c[i]]+w[i]);//这里可能 刚学的时候不理解，自己模拟一下就懂了； } } printf("%d\n",val[V]); } return 0; }