piggy bank 完全背包

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

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

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.".

完全背包且此题的背包容量必须等于物品总容量不能有空余！且是寻找最小总价值！

使用空间优化，背包容量正序遍历，这点与01背包是不同的，空间优化的01背包采用倒叙遍历。

比普通01背包初始化情况不同，INF。

进行dp时，需要额外附加一条判断条件判断历史状态是否可以获得，这是恰好装满背包问题必须要注意的一点

#include<stdio.h>
#include<algorithm>
#define INF 0x7fffffff
using namespace std;
struct coin
{int v;int w;
}list[505];
int dp[10005];
int main()
{int t;scanf("%d",&t);int e,f;int n;while(t--){scanf("%d%d",&e,&f);f-=e;//f是背包容量 scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d%d",&list[i].v,&list[i].w);}//背包必须恰好装满 初始化与普通背包初始化不同 for(int i=0;i<=f;i++){dp[i]=INF;}dp[0]=0;//尽管是完全背包 但是依然得按照从第个一物品到最后一个物品的顺序考虑添加到包里//外层循环不能少 for(int i=1;i<=n;i++){//题目要求最小钱数 所以上面是正INF 这里用min for(int j=list[i].w;j<=f;j++){//这里这句判断必不能少，少了就错 与正确答案是两种情况//这一句判断是恰好装满背包问题必须要加注意的一句代码 //如果dp[j-list[i].w]!=INF即dp[j-list[i].w]是可到达的，可以由这种情况转移而来 if(dp[j-list[i].w]!=INF)dp[j]=min(dp[j],dp[j-list[i].w]+list[i].v);}}if(dp[f]!=INF){printf("The minimum amount of money in the piggy-bank is %d.\n",dp[f]);}else{printf("This is impossible.\n");}}  return 0;
}

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piggy bank 完全背包

题目描述

输入

输出

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