cf1342 D. Multiple Testcases

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题解

这题我感觉不是很好想，可能最近算法题做多了智商下降？

我的做法是二分答案

然后贪心分配，假设现在有$tot$组数据，那我就按照测试样例从大到小的顺序依次装入每组数据，装的时候就$1,2,3,\dots,tot, 1,2,3,\dots,tot$这样循环放

至于为什么呢，我感觉就是种贪心直觉吧

代码

#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;
}