CF1519F - Chests and Keys
CF1519F - Chests and Keys
题目大意
给定\(n\)个宝箱,\(m\)种锁和对应的钥匙
每个箱子有\(a_i\)块钱,每种钥匙\(b_i\)块,给箱子\(i\)装上\(j\)这种锁需要\(c_{i,j}\)的代价
一个箱子可以装多把锁
求最小的代价,使得无论怎么买钥匙取开箱子都无法赚钱
\(n,m\leq 6,a_i,b_i\leq 4\)
??????
《关于Codeforce 3200 是朴素搜索一事》
复杂度上限?大概\(2^{41}++\)
再把-1判掉
->31ms
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a,i##end=b;i<=i##end;++i)
#define drep(i,a,b) for(int i=a,i##end=b;i>=i##end;--i)
const int N=10,P=998244353;
int n,m;
int a[N],b[N],c[N][N];
int A[1<<6],B[1<<6],s[N][1<<6],bin[1<<6];
int C[1<<6];
int st[N];
int ans=1e9;
void dfs(int p,int s){
if(s>=ans) return;
if(p==n) { ans=s; return; }
rep(S,0,(1<<m)-1) {
int fl=1;
if(s+::s[p][S]>=ans) continue;
st[p]=S;
drep(T,(1<<(p+1))-1,1<<p) {
C[T]=C[T^(1<<p)]|st[p];
if(A[T]>B[C[T]]) {
fl=0;
break;
}
}
if(fl) dfs(p+1,s+::s[p][S]);
}
}
int main(){
scanf("%d%d",&n,&m);
rep(i,0,max(n,m)-1) bin[1<<i]=i;
rep(i,0,n-1) scanf("%d",a+i);
rep(i,0,m-1) scanf("%d",b+i);
rep(i,0,n-1) rep(j,0,m-1) scanf("%d",c[i]+j);
rep(i,0,n-1) rep(j,i+1,n-1) if(a[j]>a[i]) swap(a[i],a[j]),swap(c[i],c[j]);
rep(S,1,(1<<n)-1) A[S]=A[S&(S-1)]+a[bin[S&-S]];
rep(S,1,(1<<m)-1) B[S]=B[S&(S-1)]+b[bin[S&-S]];
rep(i,0,n-1) rep(S,1,(1<<m)-1) s[i][S]=s[i][S&(S-1)]+c[i][bin[S&-S]];
int s=0;
rep(i,0,n-1) s+=a[i];
rep(i,0,m-1) s-=b[i];
if(s>0) return puts("-1"),0;
dfs(0,0);
printf("%d\n",ans==1e9?-1:ans);
}