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

1
5 10
1 2 3 4 5
5 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 1005

int 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
[v]);                          //最优解
}
return 0;
}