HDU 2602 Bone Collector 收集骨头+最基本的01背包

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HDU 2602 Bone Collector

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Appoint description:

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解题思路：刚看过01背包，感觉还可以题意：一个人收集骨头，下面有5个骨头他背了一个容量为10的包第一行为骨头的价值第二行为骨头的体积
要你求包里面能装下的最多的骨头的价值
就是一个状态转换方程要把握好dp[j] = max(dp[j],dp[j-volume[i]]+value[i]) ;
其实他是由dp[i][j] = max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);
转换而来的，但是考虑到之前的值没有用到，所以可以变成一维数组，通过刷新来实现dp


///01背包基础#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1005 ;int value[maxn] ;int volume[maxn] ;int dp[maxn] ;int N,V ;///物品数目和背包容量int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&N,&V);        int n=N ;        for(int i=1;i<=n;i++)scanf("%d",&value[i]);        for(int i=1;i<=n;i++)scanf("%d",&volume[i]);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++){///第i个物品放入容量为v的背包获得的最大价值            for(int j=V;j>=volume[i];j--){                dp[j] = max(dp[j],dp[j-volume[i]]+value[i]) ;            }        }        printf("%d\n",dp[V]) ;    }    return 0;}/*15 101 2 3 4 55 4 3 2 1样例数5件物品背包容量为10第一行为价值第二行为体积*/

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