F - Piggy-Bank HDU 1114 （完全背包的变形+初始化细节）

F - Piggy-Bank
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 1114

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.

普通的完全背包是，求价值最大，这里正好是求价值最小，只需要把max改为min，同时由于是恰好装满，所以初始化dp数组注意下即可

dp[i]:表示到达状态i获得的最小的钱数

#include<bits/stdc++.h>using namespace std;template<class T>inline T read(T&x){    char c;    while((c=getchar())<=32)if(c==EOF)return 0;    bool ok=false;    if(c=='-')ok=true,c=getchar();    for(x=0; c>32; c=getchar())        x=x*10+c-'0';    if(ok)x=-x;    return 1;}template<class T> inline T read_(T&x,T&y){    return read(x)&&read(y);}template<class T> inline T read__(T&x,T&y,T&z){    return read(x)&&read(y)&&read(z);}template<class T> inline void write(T x){    if(x<0)putchar('-'),x=-x;    if(x<10)putchar(x+'0');    else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){    write(x);    putchar('\n');}//-------ZCC IO template------const int maxn=1e6+1000;const double inf=999999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,t,n) for(int i=(t);i<(n);i++)typedef long long  LL;typedef double DB;typedef pair<int,int> P;#define bug printf("---\n");#define mod  100007int dp[maxn];int main(){    int n,E,F;    int T;    read(T);    while(T--)    {        read__(E,F,n);        int m=F-E;        dp[0]=0;//填满体积为0 的背包，价值是0        fill(dp+1,dp+m+1,inf);//其他的都是还没有填的，初始化inf        for(int i=0;i<n;i++)        {            int p,w;            read_(p,w);            for(int j=w;j<=m;j++)                dp[j]=min(dp[j-w]+p,dp[j]);//和完全背包只是max与min的区别        }        if(dp[m]!=inf)            printf("The minimum amount of money in the piggy-bank is %d.\n",dp[m]);        else            printf("This is impossible.\n");    }    return 0;}