C - Piggy-Bank HDU - 1114

标签：else   背包   include   wan   algorithm   acm   --   sts   can   

InputThe 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.
OutputPrint 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.

题目描述：

给出存钱罐空的时候的重量和装完硬币后的重量。再给出n种不同的硬币。每种有对应的价值p和重量w。问对应重量的硬币总价值最小试多少。没有就输出不可能。

分析：

硬币的质量不限，每种可以选任意件。就是完全背包问题。假设总质量为W。只要能够推到W，就说明有解。

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const ll INF=9e15;
int p[50004];
int w[10004];
ll dp[10006];
int main(void)
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(p,0,sizeof(p));
        memset(w,0,sizeof(w));
        int e,f;
        scanf("%d%d",&e,&f);
        int n;
        cin>>n;
        int W=f-e;
        for(int i=0;i<=W;i++)
        {
            dp[i]=INF;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&p[i],&w[i]);
        }
        dp[0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=w[i];j<=W;j++)
            {
                dp[j]=min(dp[j],dp[j-w[i]]+p[i]);
            }
        }
        if(dp[W]!=INF) 
        printf("The minimum amount of money in the piggy-bank is %lld.\n",dp[W]);
        else
        printf("This is impossible.\n");
    }
    return 0;
} 
?

C - Piggy-Bank HDU - 1114

标签：else   背包   include   wan   algorithm   acm   --   sts   can   

原文地址：https://www.cnblogs.com/studyshare777/p/12307714.html