0/1背包问题

Smiling & Weeping

　　　　　　　　　　　　　　　　----就算将仙人掌的刺全部拔光，也成为不了你想要的花

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

详细判题通道：

Problem - 2602 (hdu.edu.cn)

对于这道题目的理解：

　　1.DP状态的设计：dp[i][j]代表前i个装进容量为j的背包

　　2.DP状态的转移：（1）.如果bone[i]的体积大于背包容量，直接dp[i][j] = dp[i-1][j]

　　（2）.如果bone[i][j]的体积小于背包容量，判断所承受的最大价值

　　dp[i][j] = max(dp[i-1][j-bone[i].volume] + bone[i].value , dp[i-1][j])

 

#include<iostream>
#include<stdlib.h>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long ll;
ll dp[1010][1010];     //把前i个bone放入空间为j的背包
struct data
{
    ll value , volume;
}bone[10000];
int main()
{
    int T , N , V;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp , 0  , sizeof(dp));
        scanf("%d%d",&N,&V);
        for(int i = 1; i <= N; i++) scanf("%lld",&bone[i].value);
        for(int i = 1; i <= N; i++) scanf("%lld",&bone[i].volume);
        for(int i = 1; i <= N; i++)
            for(int j = 0; j <= V; j++)　　//注意背包体积要从0开始
            {
                if(bone[i].volume 》 j) dp[i][j] = dp[i-1][j];
                else
                    dp[i][j] = max(dp[i-1][j-bone[i].volume] + bone[i].value, dp[i-1][j]);
            }
        printf("%lld\n",dp[N][V]);
    }
    return 0;
}