[luogu4026 SHOI2008]循环的债务 (DP)

传送门

吐槽洛谷难度标签qwq

Solution

显然是一道神奇的DP，由于总钱数不变，我们只需要枚举前两个人的钱数就可知第三个人的钱数

DP的时候先枚举只用前k个币种，然后枚举前两个人的钱数，然后枚举转移即可

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
using namespace std;

inline int read() {
	int x=0,f=1;char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int N=1010,INF=0x3f3f3f3f;
int p[7]={0,100,50,20,10,5,1};
int sum,s1,s2,x1,x2,x3,ans;
int f[7][N][N],num[4][7];

int main() {
	x1=read(),x2=read(),x3=read();
	F(i,1,3) F(j,1,6) sum+=(num[i][j]=read())*p[j];
	F(i,1,6) s1+=num[1][i]*p[i],s2+=num[2][i]*p[i];
	if(s1-x1+x3>sum||s2-x2+x1>sum||sum-s1-s2-x3+x2>sum)
		return puts("impossible"),0;
	memset(f,0x3f,sizeof(f));f[0][s1][s2]=0;
	F(k,1,6) {
		int tot=num[1][k]+num[2][k]+num[3][k];
		F(i,0,sum) F(j,0,sum-i) {
			if(f[k-1][i][j]==INF) continue;
			F(x,0,tot) F(y,0,tot-x) {
				int z=tot-x-y;
				int s=(abs(num[1][k]-x)+abs(num[2][k]-y)+abs(num[3][k]-z))/2;
				int d1=i-(num[1][k]-x)*p[k],d2=j-(num[2][k]-y)*p[k];
				if(d1+d2>sum) continue; if(d1<0||d2<0) continue;
				f[k][d1][d2]=min(f[k][d1][d2],f[k-1][i][j]+s);
			}
		}
	}
	int ans=f[6][s1-x1+x3][s2-x2+x1];
	if(ans==INF) return puts("impossible"),0;
	return printf("%d",ans),0;
}