Piggy-Bank
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 557 Accepted Submission(s): 304 |
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Problem Description Before ACM can do anything, a budget must be prepared and the But there is |
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Input The input consists of T test cases. The number of them (T) is |
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Output Print exactly one line of output for each test case. The line |
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Sample Input 3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4 |
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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. |
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Source Central Europe 1999 |
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Recommend Eddy |
题目大意:
多种硬币放入一个存钱罐中(不限数目), 每种硬币 i 有价值 Vi 与重量 Wi 两个属性, 求给定重量 y-x 下最小的硬币钱数n
解题思路:
这是一个多重背包的变形[每种硬币最多有 (y-x)/Wi 个]. 但这道题目求的是在完全装满背包的前提下最小的价值.
自然想到的是将经典的背包问题的状态转移方程进行修改, 将max改为min, 但这样处理的话, 初始化的时候需要动一下脑筋
考虑正常的背包问题, 若要求的是恰好装满的状态是, 需要将除 dp[0] 之外的其他元素全部初始化为 负无穷, 最后的结果若为负无穷,说明无解(负无穷+有限数仍为负无穷, 这样就排除掉了不恰好装满的状态) .
但这道题目要求的是最小值, 所以我们需要将需 dp[0] 之外的其他元素全部初始化为 正无穷 , dp[0]初始化为0
如此即可AC
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define INF 0x7ffffff
#define MAXN 10000
using namespace std;
int dp[10010];
int main()
{
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int t;
int cc;
int e,f;
int n;
cin>>t;
int p,w;
while(t--){
cin>>e>>f;
cc=f-e;
cin>>n;
dp[0]=0;
for(int i=1;i<=cc;i++){
dp[i]=INF;
}
for(int i=0;i<n;i++){
cin>>p>>w;
for(int j=w;j<=cc;j++){
dp[j]=min(dp[j],dp[j-w]+p);
}
}
if(dp[cc]==INF){
cout<<"This is impossible."<<endl;
}
else cout<<"The minimum amount of money in the piggy-bank is "<<dp[cc]<<"."<<endl;
}
}