hdu 1114 Piggy Bank (完全背包 要求正好装满)

题目链接：hdu 1114

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description
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!

Input
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.

Output
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.”.

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

Sample Output
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.

题有点长，大意是说给T组数据，每组数据的开始是E和F两个数，分别代表存钱罐空时和满时的重量，接下来告诉你有N种硬币，以及每种硬币的P（价值）和W（重量），要求装入总价值最少的硬币使存钱罐正好能够达到F的重量。

刚开始接触背包时，做的背包题目都是要求背包装的价值尽可能大，并没有要求正好完全装满。而要把背包正好装满，也并不需要太多改动，只需在初始化的时候只把dp[0]初始化为0，其余都设置为未定义，无合法解的状态即可。

对应到程序中可以初始化为-INF，这道题要求的是最小价值，因此初始化为INF。

具体原因参照《背包九讲》关于初始化的细节问题。

F - E为背包最大容限，因为初始化时设定的是dp[0] = 0，所以起始点必须是0，最大为F - E，这里是一个小细节。

#include <cstdio>#include <cstring>#include <algorithm>#define MAX 10010#define MM 505#define INF 0x3f3f3f3fusing namespace std;int w[MM];int c[MM];int dp[MAX];int main(){    int T;    while(~scanf("%d", &T) && T >= 0)    {        while(T--)        {            int E, F;            scanf("%d %d", &E, &F);            int N;            scanf("%d", &N);            for(int i = 0; i < N; i++)            {                scanf("%d %d", &w[i], &c[i]);            }            memset(dp, INF, sizeof(dp));            dp[0] = 0;            for(int i = 0; i < N; i++)            {                for(int j = c[i]; j <= F - E; j++)                {                    dp[j] = min(dp[j], dp[j - c[i]] + w[i]);                }            }            if(dp[F - E] == INF)                printf("This is impossible.\n");            else                printf("The minimum amount of money in the piggy-bank is %d.\n", dp[F - E]);        }    }    return 0;}

运行结果：