如题。$N \leqslant 5000$。


感觉自己思路永远都是弯弯绕绕的。。即使会做也会被做繁掉。。果然还是我太菜了。


递减不爽,先倒序输入算了。第一问做个LIS没什么说的。第二问统计个数,考虑什么时候是重复计算的。$g[i]$表示第$i$个数结尾的LIS长度的方案(不重复)。当统计时dp到一个数时显然从前面满足$f_j+1=f_i且A_j<A_i$条件的累加过来,$A_j$不同的时候,自然不会有重复;当有相同的数且f一样时,如果这几种都加入,就重复了。而相同的几个数字显然靠后的方案统计到的更多,所以每次只取最靠右的那个数累加上去即可。实现上,开一个桶,记录vis,结束后再吐出来。

 #include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define dbg(x) cerr<<#x<<" = "<<x<<endl
#define _dbg(x,y) cerr<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
using namespace std;
typedef long long ll;
template<typename T>inline char MIN(T&A,T B){return A>B?A=B,:;}
template<typename T>inline char MAX(T&A,T B){return A<B?A=B,:;}
template<typename T>inline T _min(T A,T B){return A<B?A:B;}
template<typename T>inline T _max(T A,T B){return A>B?A:B;}
template<typename T>inline T read(T&x){
x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
}
const int N=+,M=<<,INF=0x3f3f3f3f;
int f[N],g[N],lis[N],a[N],vis[M],bin[N];
int n,len,ans,tot,cnt; int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
read(n);for(register int i=;i<=n;++i)read(a[n-i+]);
lis[len=]=INF;
for(register int i=;i<=n;++i){
f[i]=lower_bound(lis+,lis+len+,a[i])-lis;
if(f[i]>len)lis[++len]=a[i];else lis[f[i]]=a[i];
MAX(ans,f[i]);
}
for(register int i=;i<=n;++i){
if(f[i]==){g[i]=;continue;}
for(register int j=i-;j;--j)if(!vis[a[j]]&&a[j]<a[i]&&f[j]+==f[i])vis[bin[++tot]=a[j]]=,g[i]+=g[j];
while(tot)vis[bin[tot--]]=;
}
for(register int i=n;i;--i) if(!vis[a[i]]&&f[i]==ans)cnt+=g[i],vis[a[i]]=;
printf("%d %d\n",ans,cnt);
return ;
}

嗯这个是本人极其繁琐的想法。发现自己傻了有没有。。而且数据大的话桶不就挂了吗。。所以依据原来思路,改变对于dp状态的定义。$g[i]$表示第$i$个数结尾的长度为LIS的方案数,且不包括之前所有和自己相同且$len_{LIS}$相同的数的方案。这样每次转移时遇到相同即break。保证取的决策一定来源于上一个相同数和现在这个数之间。具体还是看code吧。。

另外还有一种做法是网络流。。不想写了QwQ。  ←不行,建边都会建炸掉

 #include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define dbg(x) cerr<<#x<<" = "<<x<<endl
#define _dbg(x,y) cerr<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
using namespace std;
typedef long long ll;
template<typename T>inline char MIN(T&A,T B){return A>B?A=B,:;}
template<typename T>inline char MAX(T&A,T B){return A<B?A=B,:;}
template<typename T>inline T _min(T A,T B){return A<B?A:B;}
template<typename T>inline T _max(T A,T B){return A>B?A:B;}
template<typename T>inline T read(T&x){
x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
}
const int N=+,M=<<,INF=0x3f3f3f3f;
int f[N],g[N],lis[N],a[N];
int n,len,ans,tot,cnt; int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
read(n);for(register int i=;i<=n;++i)read(a[n-i+]);
lis[len=]=INF;
for(register int i=;i<=n;++i){
f[i]=lower_bound(lis+,lis+len+,a[i])-lis;
if(f[i]>len)lis[++len]=a[i];else lis[f[i]]=a[i];
MAX(ans,f[i]);
}
g[]=;
for(register int i=;i<=n;++i){
for(register int j=i-;~j;--j){
if(a[i]==a[j]&&f[i]==f[j])break;
if(f[i]==f[j]+&&a[j]<a[i])g[i]+=g[j];
}
}
for(register int i=;i<=n;++i)if(f[i]==ans)cnt+=g[i];
printf("%d %d\n",ans,cnt);
return ;
}

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