背包问题（2）01背包 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22998    Accepted Submission(s): 9320

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14
思路：比上个问题更水的背包问题， 因看错了volume 和 value 而抑郁了半天（侧面说明English 
is very important！！！）， 用 ans[i][j] 表示前i个物品在容量为j的背包里的最优解， 每个问题可以分解成两个子问题的和（第i个物品放或不放）：ans[i-1][j] 前i-1个物品放入容量为j的背包里的最优解， 和ans[i-1][j - weight[i]] + value[i], 将第i个物品加入到背包内， 递推式为：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。
#include <stdio.h>#include <string.h>#include <stdlib.h>#define MAX 1005int ans[MAX][MAX], weight[MAX], value[MAX];int mymax(int a, int b){    return a > b ? a : b;}int main(){    int t, n, v, i, j;                                    //变量见题目scanf("%d", &t);while(t--){memset(ans, 0, sizeof(ans));                       //每次清空scanf("%d%d", &n, &v);for(i = 1; i <= n; i++){    scanf("%d", &value[i]);}for(i = 1; i <= n; i++){    scanf("%d", &weight[i]);}for(i = 1; i <= n; i++){    for(j = 0; j <= v; j++){if(j < weight[i])                           //在j < weight[i] 时无法加入， 保存（i-1）的解{    ans[i][j] = ans[i-1][j];}else                                         //否则取最优解（加或不加）{        ans[i][j] = mymax(ans[i-1][j], ans[i-1][j-weight[i]] + value[i]);}}}printf("%d\n", ans[n][v]);                          //最优解}    return 0;}