找零钱「Usaco2006 Dec」

题意

FJ要买一个东西，他有(n)种货币，每种货币价值为(v_i)，数量为(c_i)，而商店老板每种货币都有无限个。求交付硬币数与找零硬币数之和的最小值。

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题意

对于FJ，此题拆成01背包跑就可以过了；对于老板，跑完全背包。

代码

#include <bits/stdc++.h>

using namespace std;

namespace StandardIO {
    
    template<typename T> inline void read (T &x) {
        x=0;T f=1;char c=getchar();
        for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
        for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
        x*=f;
    }
    template<typename T> inline void write (T x) {
        if (x<0) putchar('-'),x=-x;
        if (x>=10) write(x/10);
        putchar(x%10+'0');
    }
    
}

using namespace StandardIO;

namespace Solve {
	
	const int N=101;
	const int T=10000;
	const int INF=0x3f3f3f3f;
	
	int n,t,ans=INF;
	int v[N],c[N];
	int f[T*2+1],g[T*2+1];
    
    inline void MAIN () {
    	read(n),read(t);
    	for (register int i=1; i<=n; ++i) read(v[i]);
		for (register int i=1; i<=n; ++i) read(c[i]);
		memset(f,0x3f,sizeof(f)),memset(g,0x3f,sizeof(g));
		f[0]=g[0]=0;
		for (register int i=1; i<=n; ++i) {
			for (register int j=2*T; j>=v[i]; --j) {
				for (register int k=1; k<=c[i]; ++k) {
					if (k*v[i]>j) break;
					f[j]=min(f[j],f[j-k*v[i]]+k);
				}
			}
		}
		for (register int i=1; i<=n; ++i) {
			for (register int j=v[i]; j<=2*T; ++j) {
				g[j]=min(g[j],g[j-v[i]]+1);
			}
		}
		for (register int i=t; i<=T*2; ++i) {
			if (f[i]==INF||g[i-t]==INF) continue;
			ans=min(ans,f[i]+g[i-t]);
		}
		if (ans==INF) write(-1);
		else write(ans);
    }
    
}

int main () {
    Solve::MAIN();
}