鏈接
題解
這題我感覺不是很好想,可能最近算法題做多了智商下降?
我的做法是二分答案
然後貪心分配,假設現在有組數據,那我就按照測試樣例從大到小的順序依次裝入每組數據,裝的時候就這樣循環放
至於爲什麼呢,我感覺就是種貪心直覺吧
代碼
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
ll n, m, a[maxn], b[maxn], f[maxn], cnt[maxn];
vector<ll> ans[maxn];
bool check(ll tot)
{
ll i, now=1;
rep(i,1,tot)cnt[i]=0;
drep(i,n,1)
{
cnt[now]++;
if(f[cnt[now]]<a[i])return false;
now++;
if(now>tot)now=1;
}
return true;
}
int main()
{
ll i, j, tot;
n=read(), m=read();
rep(i,1,n)a[i]=read();
rep(i,1,m)b[i]=read();
rep(i,1,m)f[b[i]]=i;
drep(j,n,1)if(f[j]==0)f[j]=f[j+1];
tot=1;
sort(a+1,a+n+1);
ll l=1, r=n;
while(l<r)
{
ll mid = (l+r)>>1;
if(check(mid))r=mid;
else l=mid+1;
}
printf("%lld\n",l);
drep(i,n,1)ans[i%l].emb(a[i]);
rep(i,0,l-1)
{
printf("%d ",ans[i].size());
for(auto x:ans[i])printf("%lld ",x);
putchar(10);
}
return 0;
}