HDU 2602 OJ Bone Collector

01背包模板题。就是求一定空间下的最大价值问题、

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22041    Accepted Submission(s): 8912

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

 

#include <stdio.h>int max(int x, int y){    if(x > y)    return x;    else    return y;}int dp[1010][1010];int main(){    int t, n, v, i, j;    int c[10101], w[10101];    while(~scanf("%d",&t))    {        while(t--)        {            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 = 0; i <= n; i++)                dp[i][0] = 0;            for(i = 0; i <= v; i++)                dp[0][i] = 0;            for(i = 1; i <= n; i++)            {                for(j = 0; j <= v; j++)                {                    dp[i][j] = dp[i-1][j];                    if(j >= c[i])                    dp[i][j] = max(dp[i-1][j] , dp[i-1][j-c[i]] + w[i]);                }            }            printf("%d\n",dp[n][v]);        }    }    return 0;}

#include <stdio.h>#include <iostream>using namespace std;int _max(int a, int b){    return a > b ? a : b;}int main(){    int w[10100], v[10100], dp[10100];    int i, j;    int n, m, t;    while(~scanf("%d",&t))    {        while(t--)        {            scanf("%d %d",&n, &m);            for(i = 1; i <= n; i++)                scanf("%d",&w[i]);            for(i = 1; i <= n; i++)                scanf("%d",&v[i]);            for(i = 0; i <= m; i++)                dp[i] = 0;            for(i = 1; i <= n; i++)            {                for(j = m; j >= v[i]; j--)                    dp[j] = _max(dp[j], dp[j-v[i]]+w[i]);            }            printf("%d\n",dp[m]);        }    }    return 0;}