01背包，完全背包， 多重背包

 多重背包伪代码如下：

 for i=1..N

    for v=0..V

        f[v]=max{f[v],f[v-weight[i]]+value[i]}

注意是顺序结构的；

01背包伪代码：

for i=1..N

    for v=V..0

        f[v]=max{f[v],f[v-weight[i]]+value[i]};

注意是逆序结构的；其中 v 代表当前背包的重，也就是说，v是从weight[i]或者是volumn[i]开始的

例子就拿 HDU 2602 Bone collector ------------01背包题型

/*
title: HDU 2602 Bone Collector
URL:http://acm.hdu.edu.cn/showproblem.php?pid=2602
by  AnthonyDU;
time: 2011, 24/4;
Problem : 2602 ( Bone Collector )     Judge Status : Accepted
RunId : 3859571    Language : C++    Author : a312745658
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

*/
#include <iostream>
using namespace std;
int value[1001], volumn[1001], dp[1001];

int max( int x, int y )
{
	return x > y ? x : y;
}

int main()
{
	int Max, i, j, k;
	int Tcase, num, vol;
	cin >> Tcase;
	while( Tcase -- )
	{
		cin >> num >> vol;
		for( i = 0; i < num; i ++ )
			cin >> value[i];
		for( i = 0; i < num; i++ )
			cin >> volumn[i];
		memset( dp , 0, sizeof( dp ));
		for( i = 0; i < num; i ++)
			for( j = vol; j >= volumn[i] ; j --)
				dp[j] = max( dp[j - volumn[i]]  + value[i], dp[j] );
		cout << dp[vol] << endl;
	}
	return 0;
}
 

以及 HDU 1114 piggy-Bank

/*
title : Piggy-Bank
URL : http://acm.hdu.edu.cn/showproblem.php?pid=1114
by AnthoyDu

Problem : 1114 ( Piggy-Bank )     Judge Status : Accepted
RunId : 3862008    Language : C++    Author : a312745658
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

*/
#include<iostream>
#include<string.h>
using namespace std;
#define INF 1000000001

int max( int x, int y)
{
    return x < y ? x : y ;
}
int dp[10005], value[502], weight[502];
int main()
{
    int Tcase, E, F, num;
    int i, j, k, Max;
    cin >> Tcase;
    while( Tcase -- )
    {
        cin >> E >> F;
        cin >> num;
        for ( i = 0; i < num; i ++ )
            cin >> value[i] >> weight[i];
        for( i = 1; i <= F - E; i ++ )
            dp[i] = INF;
        dp[0] = 0;
        for( i = 0; i < num; i ++ )
            for ( j = weight[i]; j <= F - E ;j ++ )
            {
                dp[j] = max( dp[j], dp[j-weight[i]]+ value[i]);
            }
        if( dp[F-E] == INF)
            cout << "This is impossible." << endl;
        else
            cout << "The minimum amount of money in the piggy-bank is " << dp[F-E] << "." << endl;
    }

} 

再晒一题，因为现在貌似把多重背包弄懂了。

/* title:悼念512汶川大地震遇难同胞——珍惜现在，感恩生活 Url: http://acm.hdu.edu.cn/showproblem.php?pid=2191 -------------------------- by AnthonyDu ------------------------- 题型：多重背包 Problem : 2191 ( 悼念512汶川大地震遇难同胞——珍惜现在，感恩生活 ) Judge Status : Accepted RunId : 3862308 Language : C++ Author : a312745658 */ #include<iostream> #include<string.h> using namespace std; int MAX( int x, int y ) { return x > y ? x : y; } int dp[10005], value[102], weight[102], number[102]; int main() { int Tcase, money, kind; int i, j, k, max; cin >> Tcase; while( Tcase -- ) { memset( dp, 0, sizeof( dp ) ); cin >> money >> kind; for( i = 0; i < kind; i ++ ) cin >> value[i] >> weight[i] >> number[i]; for( i = 0; i < kind; i ++ ) for( j = 0; j < number[i]; j ++ ) //与完全背包相比，就只有多一步这个，就是枚举每类大米有限袋的情况 for ( k = money; k >= value[i]; k -- ) dp[k] = MAX( dp[k], dp[k - value[i]] + weight[i]); cout << dp[money] << endl; } return 0; }