dp 完全背包

/*
题目1454：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：
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.".
样例输入：
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
样例输出：
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.
*/
//完全背包：有一个容积为V的背包，同时有n个物品，每个物品均有各自的体积w和
//价值v，每个物品的数量均为无限个，求使用该背包最多能装的物品价值总和
//完全背包 无穷个 装满
#include <stdio.h>
#define INF 0x7fffffff
int min(int a ,int b){return a < b ? a : b; }//取最小值函数
struct E{//代表钱币结构体
int w;//重量
int v;//价值
}list[501];
int dp[10001];//状态
int main(){
  int T;
  scanf("%d",&T);//输入测试数据组数
  while(T--){//T次循环，处理T组数据
 int s,tmp,i,j;
 scanf("%d%d",&tmp,&s);//输入空存储罐重量和装满钱币的存储罐重量
  s -= tmp;//计算钱币所占重量
  int n;
  scanf("%d",&n);
  for(i = 1;i<=n;i++){
   scanf("%d%d",&list[i].v,&list[i].w);
}//输入数据
for(i = 0;i<=s;i++){
   dp[i] = INF;
}
   dp[0] = 0;//因为要求所有物品恰好装满，所以初始时，除dp[0]外，其余dp[j]均为无穷
   for(i = 1;i<=n;i++){//遍历所有物品
    for(j = list[i].w;j<=s;j++){//完全背包，顺序遍历所有可能转移的状态
     if(dp[j-list[i].w] != INF)//若dp[j-list[i].w]不为无穷，就可以由此状态转移而来
       dp[j] = min(dp[j],dp[j-list[i].w] + list[i].v);//取转移值和原值的较小值
}
}
    if(dp[s] != INF)//若存在一种方案使背包恰好装满，输出其最小值 
      printf("The minimum amount of money in the piggy-bank is %d.\n",dp[s]);
    else //若不存在方案
      puts("This is impossible.");
} 
  return 0;
}
/*
0-1背包，每个物品可选数量为0或者1，完全背包，每个物品可选的数量为无穷
，两者解法基本相同，仅为状态更新时的遍历顺序，时空复杂度基本一致。
*/
/*
多重背包，介于两者之间，有容积为v的背包，给定一些物品，每种物品包含体积w，
价值v和数量k，求用该背包能装下的最大价值总量。可选物品数量为k
*/