Piggy-Bank(完全背包)

发布于 2020-10-18  0 次阅读


题目大意:

Piggy-Bank
给你一个存钱罐的初始重量和装满后的重量
给你n个可以任意取的硬币,每个硬币具有价值和重量,求将存钱罐装满后硬币的最小价值
若不能装满则输出 This is impossible.

思路:

很明显的完全背包问题,因为求的最小值,所以将数组初始化的时候为无穷大,最后判断是否装满即为装满时大小是否为无穷大即可

AC Code:

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<math.h>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
// #define int long long
#define debug(a) cout<<#a<<"="<<a<<endl;
// #define TDS_ACM_LOCAL
typedef long long ll;
const double PI=acos(-1.0);
const double e=exp(1.0);
const int M=1e9 +7;
const int N=1e5 +9;
inline int mymax(int x,int y){return x>y?x:y;}
inline int mymin(int x,int y){return x<y?x:y;}
inline int read(){
    register int x=0, f=0;
    register char ch=getchar();
    while(ch<'0'||ch>'9') f|=ch=='-',ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return f?-x:x;
}
inline void write(register int x){     
    if(x < 0) {putchar('-');x = -x;}        
    if(x > 9)  write(x/10);
    putchar(x % 10 + '0');
}
int k1, k, n;
int wei[N], v[N], dp[N];
void solve(){
    k1=read(); k=read();
    k-=k1;
    n=read();
    for(int i=1; i<=n; i++) v[i]=read(), wei[i]=read();
    memset(dp, INF, sizeof(dp));
    dp[0]=0;
    for(int i=1; i<=n; i++)
        for(int j=wei[i]; j<=k; j++)
            dp[j]=min(dp[j], dp[j-wei[i]]+v[i]);
    if(dp[k]!=INF)
        cout<<"The minimum amount of money in the piggy-bank is "<<dp[k]<<".\n";
    else 
        cout<<"This is impossible.\n";
	return ;
}

signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
    freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
    freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
    int T;
    T=read();
    while(T--) solve();
    return 0;
}


平平无奇的在校大学生