HDU2602 Bone Collector （简单01背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21250    Accepted Submission(s): 8521

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

是一道 简单的01背包问题，分别采用了二维数组和一维数组来写，本题有个很变态的数据，就是有价值没有体积的bone，尼玛这不科学！

# include<stdio.h># include<string.h>const int MAXN = 1000+5;int f[MAXN][MAXN];int knapsack(int n, int m){    int i, j, c[MAXN], w[MAXN];    for(i=1; i<=n; i++)       scanf("%d", &w[i]);for(i=1; i<=n; i++)   scanf("%d", &c[i]);    memset(f, 0, sizeof(f));    for(i=1; i<=n; i++)       for(j=0; j<=m; j++)       {           if(c[i]<=j)            {               if(w[i]+f[i-1][j-c[i]]>f[i-1][j])                 f[i][j] = w[i]+f[i-1][j-c[i]];               else                 f[i][j] = f[i-1][j];           }           else               f[i][j] = f[i-1][j];       }  return (f[n][m]);}int main(void){    int m, n, i, j, t;scanf("%d", &t);while(t--){    scanf("%d%d", &n, &m);    printf("%d\n", knapsack(n, m));}return 0;}

# include<stdio.h># include<string.h>const int MAXN = 1000+5;int f[MAXN];void knapsack(int n, int m){    int i, j;    int c[MAXN], w[MAXN];    memset(f,0,sizeof(f));    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=m; j>=0; j--) //可能存着有价值无体积的数据       {             if(j-c[i]>=0)             {                 if(f[j-c[i]]+w[i]>f[j])                 f[j] = f[j-c[i]]+w[i];             }       }    printf("%d\n", f[m]);}int main(void){    int t, n, m;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        knapsack(n, m);    }    return 0;}