kuangbin专题十二 HDU1114 Piggy-Bank （完全背包）

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35043    Accepted Submission(s): 17420

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.

题目大意：给出存钱罐的容量，和各种货币的价值的重量，然后求存钱罐里放满的最坏情况（里面钱最少）。

思路：乍一看就是完全背包，但是用贪心一发WA，然后就学习了完全背包。注意一点，这里是求最小的价值。所以初始化条件要注意一下。

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <algorithm>
 #include <sstream>
 #include <stack>
 using namespace std;
 #define mem(a,b) memset((a),(b),sizeof(a))
 #define mp make_pair
 #define pb push_back
 #define fi first
 #define se second
 #define sz(x) (int)x.size()
 #define all(x) x.begin(),x.end()
 typedef long long ll;
 const int inf = 0x3f3f3f3f;
 const ll INF =0x3f3f3f3f3f3f3f3f;
 const double pi = acos(-1.0);
 const double eps = 1e-;
 const ll mod = 1e9+;
 //head

 const int MAX =  + ;
 int w[MAX], c[MAX];
 int dp[+]; //dp[i][v] 表示 前i个物品恰好放到容量为v的 最小价值

 int main() {
     int _, n, st, ed;
     for(scanf("%d", &_);_;_--) {
         scanf("%d%d", &st, &ed);
         scanf("%d", &n);
         for(int i = ; i < n; i++) {
             scanf("%d%d", &c[i], &w[i]);
         }
         int V = ed - st;//容量
         for(int i = ; i <= V; i++)
             dp[i] = inf;// 初始化位inf 如果dp[V]为inf 那么就是没有装满
         dp[] = ;//初始化~！！
         for(int i = ; i < n; i++) {//核心代码
             for(int v = w[i]; v <= V; v++)
                 dp[v] = min(dp[v], dp[v-w[i]] + c[i]);
         }
         if(dp[V] != inf)
             printf("The minimum amount of money in the piggy-bank is %d.\n", dp[V]);
         else
             printf("This is impossible.\n");
     }
 }