背包问题之完全背包

     完全背包与0-1背包之间区别是在完全背包中，每个物品可以无限次选择。代码上的区别主要是把逆序更新改为了顺序更新。

     0-1背包介绍见另一随笔http://www.cnblogs.com/yuxiaoba/p/8458135.html

Piggy-Bank

题目描述

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid. 

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs! 

输入描述:

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams. 
       

输出描述:

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.". 
       

示例1

输入

3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4

输出

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.

题目大意：有一个存钱罐，告知其空时的重量和当前的重量，并给定一些钱币的价值和相应的重量，求存钱罐中最少有多少现金。
由于每个钱币的数量可以任意多，所以该问题为完全背包问题。此外这里还有两处变化。
一是要求的不是最大值而是最小值，因而在状态转移时要选择较小的转移值；
二是要求恰好装满，解决这个问题时需把dp[0]置0，dp[j]置无穷，其余代码基本不变

#include <stdio.h>
#include <limits.h>

int min( int a,int b)
{
    return a < b ? a:b;
}

struct E
{
    int w;  //物品时间
    int v;  //物品价值
} list[501];
int dp[10001];

int main()
{
    int T,n;
    int k,s;  //空和满的存钱罐的重量
    int i,j;
    scanf("%d",&T);  //读入案例次数
    while( T-- )
    {
        scanf("%d%d",&k,&s);
        s -= k;  //计算钱币的重量；
        scanf("%d",&n);
        for( i=1; i<=n; i++)
            scanf("%d%d",&list[i].v,&list[i].w);
        for( i=1; i<=s; i++)
            dp[i] = INT_MAX;
        dp[0] = 0;
        for( i=1; i<=n; i++)
        {
            for( j=list[i].w; j<=s; j++)
            {
                if(dp[j-list[i].w]!=INT_MAX )
                    dp[j] = min( dp[j],dp[j-list[i].w]+list[i].v);
            }
        }
        if( dp[s]!=INT_MAX )
            printf("The minimum amount of money in the piggy-bank is %d.
",dp[s]);
        else printf("This is impossible.
");
    }

    return 0;
}