03_01背包问题

来源：刘汝佳《算法竞赛入门经典--训练指南》 P60 问题4：

问题描述：有n种物品，每种只有一个，第i种物品的体积为Vi，重量为Wi。选一些物品装到一个容量为C的背包中，使得背包内物品在总体积不超过C的前提下重量尽量大。其中，1<=n<=100。1<=V<=C<=10000。1<=Wi<=10^6。

分析：用f[i][j]表示“把前i个物品装到容量为j的背包中的最大总重量”，则状态转移方程为f[i][j] = Max{f[i-1][j],f[i-1][j-Vi]+Wi | Vi<=j};

滚动数组优化后代码：

memset(f,0,sizeof(f));
for(int i=1; i<=n; i++)
{
    scanf("%d %d",&V,&W);
    for(int j=C; j>=0; j--)
    {
        if(j>=V) 
            f[j] = Max(f[j],f[j-V]+W);
    }
}

例题来源：http://acm.hdu.edu.cn/showproblem.php?pid=2602

例题：hdu 2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：给一个体积为V( V <= 1000)的背包和n(n <= 1000 )个骨头，每个骨头i有各自的体积volume[i]和价格value[i]，求在不超过背包体积V的前提下往背包里装骨头的最大价值。

代码实现：

#include "stdio.h"
#include "string.h"
#define N 1005
int dp[N];
int value[N];  //价值数组
int volume[N]; //体积数组

int Max(int a,int b) { return a>b?a:b; }

int main()
{
    int T;
    int n,V;  //n为骨头数量，V为背包体积
    int i,j,v,w;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&V);
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            scanf("%d",&value[i]);
        for(i=0; i<n; i++)
            scanf("%d",&volume[i]);
        for(i=0; i<n; i++)
        {
            v = volume[i]; //第i个骨头的体积
            w = value[i]; //第i个骨头的价值
            for(j=V; j>=v; j--)
                dp[j] = Max(dp[j],dp[j-v]+w);
        }
        printf("%d\n",dp[V]);
    }
    return 0;
}